OP Malhotra Class 10 Maths Solutions Chapter 17 Heights and Distances Exercise 17

S Chand Class 10 ICSE Maths Solutions Chapter 17 Heights and Distances Ex 17

Question 1. The angle of elevation of the top of a tower from a point at a distance of 100 metres from its foot on a horizontal plane is found to be 60°. Find the height of the tower.
Answer: Let AB be the tower and O be the point on the ground. The angle of elevation from O to the top of the tower (A) is \( 60^\circ \). The distance from the foot of the tower (B) to point O is 100 m. We need to find the height of the tower, which is AB (let's call it \( h \)).
We can use the tangent function in the right-angled triangle AOB:
\( \tan \theta = \frac{\text{AB}}{\text{OB}} \)
\( \implies \tan 60^\circ = \frac{h}{100} \)
\( \implies h = 100 \times \tan 60^\circ \)
We know that \( \tan 60^\circ = \sqrt{3} \), and \( \sqrt{3} \approx 1.732 \).
\( \implies h = 100 \times \sqrt{3} \)
\( \implies h = 100 \times 1.732 \)
\( \implies h = 173.2 \text{ m} \)
So, the height of the tower is 173.2 metres. This calculation helps us find unknown heights or distances using angles.
In simple words: Imagine a tower and a point 100 metres away on the ground. If you look up at the top of the tower from that point, the angle is 60 degrees. To find how tall the tower is, we use a special math rule called tangent. We multiply 100 by the value of tangent 60 degrees, which is about 1.732. This gives us the tower's height as 173.2 metres.

🎯 Exam Tip: Remember to identify the right-angled triangle and correctly apply trigonometric ratios (sine, cosine, or tangent) based on the given sides and angles. Always draw a clear diagram to visualize the problem.

Question 2. A vertical flagstaff stands on a horizontal plane. From a point distant 150 metres from its foot the angle of elevation of its top is found to be 30°; find the height of the flagstaff.
Answer: Let AB be the flagstaff and O be the point on the ground. The flagstaff stands vertically, so \( \angle B = 90^\circ \). The angle of elevation from O to the top of the flagstaff (A) is \( 30^\circ \). The distance from the foot of the flagstaff (B) to point O is 150 m. We need to find the height of the flagstaff, which is AB (let's call it \( h \)).
In the right-angled triangle AOB:
\( \tan \theta = \frac{\text{AB}}{\text{OB}} \)
\( \implies \tan 30^\circ = \frac{h}{150} \)
\( \implies h = 150 \times \tan 30^\circ \)
We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \). To simplify, we can multiply the numerator and denominator by \( \sqrt{3} \).
\( \implies h = 150 \times \frac{1}{\sqrt{3}} = \frac{150 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{150 \sqrt{3}}{3} \)
\( \implies h = 50 \sqrt{3} \)
Using \( \sqrt{3} \approx 1.732 \):
\( \implies h = 50 \times 1.732 \)
\( \implies h = 86.6 \text{ m} \)
Therefore, the height of the flagstaff is 86.6 metres. Understanding basic trigonometry helps solve such real-world height problems.
In simple words: A pole, called a flagstaff, stands straight up. From a spot 150 metres away, if you look at the top of the pole, the angle is 30 degrees. To find the pole's height, we use the tangent rule: multiply 150 by tangent 30 degrees. This gives us about 86.6 metres for the height.

🎯 Exam Tip: Always rationalize the denominator when it contains a square root. Using the approximate value of \( \sqrt{3} \) (1.732) should be done only at the final step to maintain precision.

Question 3. The string of a kite is 150 long and it makes an angle of 60° with the horizontal. Find the height of the kite from the ground.
Answer: Let A be the position of the kite and AB be its height from the ground. Let C be the point on the ground where the string is held. The length of the string AC is 150 m. The string makes an angle of \( 60^\circ \) with the horizontal ground. We need to find the height of the kite, AB (let's call it \( h \)).
In the right-angled triangle ABC:
\( \sin \theta = \frac{\text{AB}}{\text{AC}} \)
\( \implies \sin 60^\circ = \frac{h}{150} \)
\( \implies h = 150 \times \sin 60^\circ \)
We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
\( \implies h = 150 \times \frac{\sqrt{3}}{2} \)
\( \implies h = 75 \sqrt{3} \)
Using \( \sqrt{3} \approx 1.732 \):
\( \implies h = 75 \times 1.732 \)
\( \implies h = 129.9 \text{ m} \)
Therefore, the height of the kite from the ground is 129.9 metres. This demonstrates how sine is used when the hypotenuse and an angle are known.
In simple words: A kite string is 150 metres long and it makes an angle of 60 degrees with the ground. To find how high the kite is flying, we use the sine rule. We multiply the string length (150 m) by the sine of 60 degrees (which is about 0.866). This calculation shows the kite is 129.9 metres high.

🎯 Exam Tip: When given the hypotenuse and an angle, use the sine or cosine function. Sine relates the opposite side to the hypotenuse, while cosine relates the adjacent side to the hypotenuse.

Question 4. If the shadow of a tower is 30 metres when the sun's elevation is 30°, what is the length of the shadow when the sun's elevation is 60° ?
Answer: Let AB be the height of the tower (let's call it \( h \)). Let C be the point where the sun's elevation is \( 30^\circ \), and D be the point where the sun's elevation is \( 60^\circ \).
First case: When the sun's elevation is \( 30^\circ \), the shadow length BC is 30 metres.
In right-angled triangle ABC:
\( \tan 30^\circ = \frac{\text{AB}}{\text{BC}} \)
\( \implies \frac{1}{\sqrt{3}} = \frac{h}{30} \)
\( \implies h = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3} \text{ m} \)
Second case: When the sun's elevation is \( 60^\circ \), let the shadow length BD be \( x \) metres.
In right-angled triangle ABD:
\( \tan 60^\circ = \frac{\text{AB}}{\text{BD}} \)
\( \implies \sqrt{3} = \frac{h}{x} \)
Now substitute the value of \( h \) we found:
\( \implies \sqrt{3} = \frac{10\sqrt{3}}{x} \)
\( \implies x\sqrt{3} = 10\sqrt{3} \)
\( \implies x = 10 \text{ m} \)
The length of the shadow when the sun's elevation is \( 60^\circ \) is 10 metres. This problem shows how shadow length changes with the sun's angle.
In simple words: A tower has a 30-metre shadow when the sun is 30 degrees high. We first use this to find the tower's height. Then, we use the tower's height to find the new shadow length when the sun is 60 degrees high. The shadow becomes shorter, at 10 metres.

🎯 Exam Tip: When dealing with changes in the angle of elevation (or depression), it's often helpful to represent the changing distances with variables and set up two separate trigonometric equations to solve for unknowns.

Question 5. The angle of elevation of the top of a tower (which is yet incomplete) at a point 120 m from its base is 45°. How much higher should it be raised so that the elevation at the same point may become 60° ?
Answer: Let AB be the base of the tower, and B be the point 120 m from the base. Let C be the top of the incomplete tower, and D be the desired top of the complete tower. So, the distance AB = 120 m.
First, find the height of the incomplete tower, BC (let's call it \( x \)). The angle of elevation to C from A is \( 45^\circ \).
In right-angled triangle ABC:
\( \tan 45^\circ = \frac{\text{BC}}{\text{AB}} \)
\( \implies 1 = \frac{x}{120} \)
\( \implies x = 120 \text{ m} \)
Next, let the tower be raised by an additional height CD (let's call it \( h \)) so that the total height is BD = \( x+h \). The new angle of elevation to D from A is \( 60^\circ \).
In right-angled triangle ABD:
\( \tan 60^\circ = \frac{\text{BD}}{\text{AB}} \)
\( \implies \sqrt{3} = \frac{x+h}{120} \)
Substitute \( x = 120 \text{ m} \):
\( \implies \sqrt{3} = \frac{120+h}{120} \)
\( \implies 120\sqrt{3} = 120+h \)
\( \implies h = 120\sqrt{3} - 120 \)
\( \implies h = 120(\sqrt{3} - 1) \)
Using \( \sqrt{3} \approx 1.732 \):
\( \implies h = 120(1.732 - 1) \)
\( \implies h = 120(0.732) \)
\( \implies h = 87.84 \text{ m} \)
The tower should be raised by 87.84 metres. This shows how to calculate the additional height needed to change an angle of elevation.
In simple words: An unfinished tower is 120 metres away. Looking at its top, the angle is 45 degrees. We want the angle to be 60 degrees. First, we find the current height. Then, using the new angle, we find what the total height should be. The difference between the new total height and the current height tells us how much more the tower needs to be built up, which is 87.84 metres.

🎯 Exam Tip: Break down complex problems into simpler right-angled triangles. Clearly label all knowns and unknowns in your diagram to avoid confusion, especially when dealing with nested or overlapping triangles.

Question 6. Find the angle of depression from the top of a tower 140 m high of an object on the ground at a distance of 200 m from the foot of the tower.
Answer: Let AB be the tower, with height AB = 140 m. Let P be the object on the ground. The distance from the foot of the tower (B) to the object (P) is PB = 200 m. We need to find the angle of depression from the top of the tower (A) to the object (P). Let this angle be \( \theta \).
The angle of depression from A to P is equal to the angle of elevation from P to A (alternate interior angles). So, in the right-angled triangle ABP, the angle at P is \( \theta \).
\( \tan \theta = \frac{\text{AB}}{\text{PB}} \)
\( \implies \tan \theta = \frac{140}{200} \)
\( \implies \tan \theta = \frac{7}{10} = 0.7 \)
To find \( \theta \), we look up the tangent value in trigonometric tables. We find that \( \tan 35^\circ \approx 0.70021 \), which is very close to 0.7.
So, \( \theta \approx 35^\circ \).
The angle of depression is approximately \( 35^\circ \). This calculation is vital in navigation and surveying to determine vertical and horizontal positions.
In simple words: Imagine a 140-metre tall tower. There's an object on the ground 200 metres away from the tower's base. To find the angle you'd look down from the top of the tower to see the object, we use the tangent rule. We divide 140 by 200, which gives 0.7. Then we find the angle whose tangent is 0.7, which is about 35 degrees.

🎯 Exam Tip: Remember that the angle of depression from the observer's eye to an object is equal to the angle of elevation from the object to the observer's eye, due to parallel lines and alternate interior angles.

Question 7. A vertical tower is 20 m high. A man at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower?
Answer: Let AB be the tower, with height AB = 20 m. Let M be the man standing at some distance from the foot of the tower (B). Let MB be this distance (let's call it \( x \)). The angle of elevation from the man to the top of the tower (A) is \( \theta \), and we are given that \( \cos \theta = 0.53 \). We need to find the distance \( x \).
In the right-angled triangle AMB, the hypotenuse AM can be found using the Pythagorean theorem: \( \text{AM} = \sqrt{\text{AB}^2 + \text{MB}^2} = \sqrt{20^2 + x^2} = \sqrt{400 + x^2} \).
The cosine of the angle \( \theta \) is given by:
\( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\text{MB}}{\text{AM}} \)
\( \implies 0.53 = \frac{x}{\sqrt{400 + x^2}} \)
To solve for \( x \), we square both sides:
\( (0.53)^2 = \frac{x^2}{400 + x^2} \)
\( 0.2809 = \frac{x^2}{400 + x^2} \)
Now, cross-multiply:
\( 0.2809 (400 + x^2) = x^2 \)
\( 112.36 + 0.2809 x^2 = x^2 \)
Rearrange the terms to solve for \( x^2 \):
\( 112.36 = x^2 - 0.2809 x^2 \)
\( 112.36 = (1 - 0.2809) x^2 \)
\( 112.36 = 0.7191 x^2 \)
\( x^2 = \frac{112.36}{0.7191} \approx 156.25 \)
Take the square root to find \( x \):
\( x = \sqrt{156.25} = 12.5 \text{ m} \)
The man is standing 12.5 metres from the foot of the tower. Using cosine helps determine horizontal distances when other information is available.
In simple words: A tower is 20 metres high. A man is standing some distance away. He knows that the "cosine" of the angle when he looks up at the tower top is 0.53. To find how far he is standing, we use this cosine value along with the tower's height. After some calculation, we find he is 12.5 metres away.

🎯 Exam Tip: If the problem gives you a cosine or sine value instead of an angle, you might need to use the Pythagorean theorem to relate the sides before applying the trigonometric ratio.

Question 8. In the figure, it is given that AB is perp. to BD and is of length x metres. DC = 30m, ∠ADB = 30° and ∠ACB = 45°. Find x.
Answer: Let AB be the vertical length, which is \( x \) metres. BD is the horizontal base. We have two points C and D on the base. DC = 30 m. The angle of elevation to A from D is \( \angle ADB = 30^\circ \), and from C is \( \angle ACB = 45^\circ \). We need to find \( x \).
Let BC be \( y \) metres. Then BD = BC + CD = \( y + 30 \) metres.
First, consider the right-angled triangle ABC:
\( \tan 45^\circ = \frac{\text{AB}}{\text{BC}} \)
\( \implies 1 = \frac{x}{y} \)
\( \implies x = y \) (Equation 1)
Next, consider the right-angled triangle ABD:
\( \tan 30^\circ = \frac{\text{AB}}{\text{BD}} \)
\( \implies \frac{1}{\sqrt{3}} = \frac{x}{y+30} \)
Substitute \( y = x \) from Equation 1 into this equation:
\( \implies \frac{1}{\sqrt{3}} = \frac{x}{x+30} \)
Cross-multiply:
\( \implies x+30 = x\sqrt{3} \)
Rearrange terms to solve for \( x \):
\( \implies 30 = x\sqrt{3} - x \)
\( \implies 30 = x(\sqrt{3} - 1) \)
\( \implies x = \frac{30}{\sqrt{3} - 1} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{3}+1}{\sqrt{3}+1} \):
\( x = \frac{30(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{30(\sqrt{3}+1)}{3-1} = \frac{30(\sqrt{3}+1)}{2} \)
\( \implies x = 15(\sqrt{3}+1) \)
Using \( \sqrt{3} \approx 1.732 \):
\( \implies x = 15(1.732+1) = 15(2.732) \)
\( \implies x = 40.98 \text{ m} \)
Thus, the length \( x \) is 40.98 metres. This problem highlights how to solve for a common height or length shared between two triangles.
In simple words: We have a vertical line (x) and two points on the ground, C and D. The distance between C and D is 30 metres. Looking from C to the top of x, the angle is 45 degrees. From D, the angle is 30 degrees. We use these angles and distances to set up two equations and find that x is 40.98 metres.

🎯 Exam Tip: Always look for relationships between the sides of different triangles (like \( y=x \) here) to simplify the system of equations. Rationalizing the denominator is a standard mathematical practice.

Question 9. In the figure, from a boat 200 in away from a vertical cliff the angles of elevation of the top and the foot of a vertical concrete pillar at the edge of the cliff are 21° and 18° 30′. Calculate:
(i) the height of the cliff;
(ii) the height of the pillar.
Answer: Let B be the position of the boat, and A be the base of the cliff. The distance from the boat to the cliff is AB = 200 m. Let CA be the height of the cliff (let it be \( h \)) and PC be the height of the pillar (let it be \( x \)). So, the total height PA = \( h+x \).
The angle of elevation to the top of the cliff (C) is \( 18^\circ 30' \).
The angle of elevation to the top of the pillar (P) is \( 21^\circ \).
(i) To find the height of the cliff (CA):
Consider the right-angled triangle CAB:
\( \tan 18^\circ 30' = \frac{\text{CA}}{\text{AB}} \)
\( \implies \tan 18.5^\circ = \frac{h}{200} \)
From trigonometric tables, \( \tan 18.5^\circ \approx 0.33460 \).
\( \implies 0.33460 = \frac{h}{200} \)
\( \implies h = 200 \times 0.33460 \)
\( \implies h = 66.920 \text{ m} \)
The height of the cliff is 66.92 metres.
(ii) To find the height of the pillar (PC):
Consider the right-angled triangle PAB:
\( \tan 21^\circ = \frac{\text{PA}}{\text{AB}} \)
\( \implies \tan 21^\circ = \frac{h+x}{200} \)
From trigonometric tables, \( \tan 21^\circ \approx 0.38386 \).
\( \implies 0.38386 = \frac{66.920+x}{200} \)
\( \implies 200 \times 0.38386 = 66.920+x \)
\( \implies 76.772 = 66.920+x \)
\( \implies x = 76.772 - 66.920 \)
\( \implies x = 9.852 \text{ m} \)
The height of the pillar is 9.852 metres. This two-part problem shows how to find individual heights within a combined structure.
In simple words: A boat is 200 metres from a cliff. From the boat, looking at the cliff's top, the angle is 18.5 degrees. On the cliff's edge is a pillar, and looking at the pillar's top, the angle is 21 degrees. We first use the 18.5-degree angle to find the cliff's height (66.92 metres). Then, we use the 21-degree angle to find the combined height of the cliff and pillar. Subtracting the cliff's height gives us the pillar's height, which is 9.852 metres.

🎯 Exam Tip: When dealing with multiple angles of elevation to points on the same vertical line, solve for the lower height first using its corresponding angle, then use that height to find the additional height from the second angle.

Question 10. In the figure, the top of a tower is observed from two points on the same horizontal line through a point on the base of the tower. If the angles of elevation at the two points be 30° and 45°, and the distance between them is 20 m, find the height of the tower.
Answer: Let TS be the tower, with height TS = \( h \). Let B be a point on the ground with an angle of elevation of \( 45^\circ \), and A be another point such that AB = 20 m, with an angle of elevation of \( 30^\circ \). The points A, B, and S (foot of the tower) are on the same horizontal line.
Let BS be \( x \) metres.
First, consider the right-angled triangle TSB:
\( \tan 45^\circ = \frac{\text{TS}}{\text{BS}} \)
\( \implies 1 = \frac{h}{x} \)
\( \implies x = h \) (Equation 1)
Next, consider the right-angled triangle TSA. The base SA = SB + BA = \( x + 20 \).
\( \tan 30^\circ = \frac{\text{TS}}{\text{SA}} \)
\( \implies \frac{1}{\sqrt{3}} = \frac{h}{x+20} \)
Substitute \( x = h \) from Equation 1:
\( \implies \frac{1}{\sqrt{3}} = \frac{h}{h+20} \)
Cross-multiply:
\( \implies h+20 = h\sqrt{3} \)
Rearrange terms to solve for \( h \):
\( \implies 20 = h\sqrt{3} - h \)
\( \implies 20 = h(\sqrt{3} - 1) \)
\( \implies h = \frac{20}{\sqrt{3} - 1} \)
Using \( \sqrt{3} \approx 1.732 \):
\( \implies h = \frac{20}{1.732 - 1} = \frac{20}{0.732} \)
\( \implies h \approx 27.32 \text{ m} \)
The height of the tower is approximately 27.32 metres. This setup is common for finding heights when direct measurement is not possible.
In simple words: Two spots on the ground are 20 metres apart. From the spot closer to a tower, you look up at 45 degrees. From the spot further away, you look up at 30 degrees. We use these angles to create two math equations. By solving them, we find the tower's height is about 27.32 metres.

🎯 Exam Tip: When using values from trigonometric tables or approximations, perform calculations to at least one more decimal place than required in the final answer to minimize rounding errors.

Question 11. A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angles of elevation to be 30°. Find (i) the height of the tree, correct to 2 decimal places ; (ii) the width of river.
Answer: Let TR be the height of the tree (let it be \( h \)). Let R be the point on the bank directly opposite the tree. Let B be the person's initial position, and BR be the width of the river (let it be \( x \)). When the person moves 40 m away, their new position is A, so AB = 40 m. The total distance from A to the tree is AR = \( x+40 \).
(i) To find the height of the tree (TR = \( h \)):
First, consider the right-angled triangle TRB (from the initial position):
\( \tan 60^\circ = \frac{\text{TR}}{\text{BR}} \)
\( \implies \sqrt{3} = \frac{h}{x} \)
\( \implies x = \frac{h}{\sqrt{3}} \) (Equation 1)
Next, consider the right-angled triangle TRA (from the new position):
\( \tan 30^\circ = \frac{\text{TR}}{\text{AR}} \)
\( \implies \frac{1}{\sqrt{3}} = \frac{h}{x+40} \)
Substitute \( x = \frac{h}{\sqrt{3}} \) from Equation 1 into this equation:
\( \implies \frac{1}{\sqrt{3}} = \frac{h}{\frac{h}{\sqrt{3}}+40} \)
Cross-multiply:
\( \implies \frac{h}{\sqrt{3}}+40 = h\sqrt{3} \)
Rearrange terms to solve for \( h \):
\( \implies 40 = h\sqrt{3} - \frac{h}{\sqrt{3}} \)
\( \implies 40 = h \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) \)
\( \implies 40 = h \left( \frac{3-1}{\sqrt{3}} \right) \)
\( \implies 40 = h \left( \frac{2}{\sqrt{3}} \right) \)
\( \implies h = \frac{40\sqrt{3}}{2} = 20\sqrt{3} \text{ m} \)
Using \( \sqrt{3} \approx 1.732 \):
\( \implies h = 20 \times 1.732 = 34.64 \text{ m} \)
The height of the tree is 34.64 metres.
(ii) To find the width of the river (BR = \( x \)):
Using Equation 1, \( x = \frac{h}{\sqrt{3}} \):
\( \implies x = \frac{20\sqrt{3}}{\sqrt{3}} \)
\( \implies x = 20 \text{ m} \)
The width of the river is 20 metres. These calculations help determine dimensions across inaccessible areas.
In simple words: A person looks at a tree across a river, seeing an angle of 60 degrees. Then, they walk 40 metres away and the angle becomes 30 degrees. We use these two angles and the distance walked to find both the height of the tree (34.64 metres) and the width of the river (20 metres).

🎯 Exam Tip: Always set up two equations from the two different observation points and then solve them simultaneously. It's often easier to express one unknown in terms of the other and substitute.

Question 12. In the figure. the angle of elevation of the top P of a vertical tower PQ from a point X Is 60°; at a point Y. 40m vertically above X, the angle of elevatlon Li 45°. Find the distance XQ. (Give your answers to the nearest metre).
Answer: Let PQ be the tower, with height PQ = \( h \). Let X be a point on the ground, and QX be its distance from the base of the tower (let it be \( x \)). The angle of elevation from X to the top of the tower (P) is \( 60^\circ \).
A point Y is 40 m vertically above X. So, YX = 40 m. From Y, the angle of elevation to P is \( 45^\circ \). Draw a line YZ parallel to QX, such that Z is on PQ. Then YZ = QX = \( x \) and QZ = YX = 40 m. Thus, PZ = PQ - QZ = \( h - 40 \).
First, consider the right-angled triangle PQX:
\( \tan 60^\circ = \frac{\text{PQ}}{\text{QX}} \)
\( \implies \sqrt{3} = \frac{h}{x} \)
\( \implies h = x\sqrt{3} \) (Equation 1)
Next, consider the right-angled triangle PZY:
\( \tan 45^\circ = \frac{\text{PZ}}{\text{YZ}} \)
\( \implies 1 = \frac{h-40}{x} \)
\( \implies x = h-40 \) (Equation 2)
Substitute \( h = x\sqrt{3} \) from Equation 1 into Equation 2:
\( \implies x = x\sqrt{3} - 40 \)
Rearrange terms to solve for \( x \):
\( \implies 40 = x\sqrt{3} - x \)
\( \implies 40 = x(\sqrt{3} - 1) \)
\( \implies x = \frac{40}{\sqrt{3} - 1} \)
Using \( \sqrt{3} \approx 1.732 \):
\( \implies x = \frac{40}{1.732 - 1} = \frac{40}{0.732} \)
\( \implies x \approx 54.644 \text{ m} \)
Rounding to the nearest metre, the distance XQ is 55 metres. This calculation involves finding horizontal distance when observing from two different vertical points.
In simple words: From a point X on the ground, the angle to the top of a tower is 60 degrees. If you go up 40 metres to point Y, the angle to the tower top is 45 degrees. We use these angles and distances to find the distance from point X to the tower's base. The answer, rounded to the nearest metre, is 55 metres.

🎯 Exam Tip: When an observation point moves vertically, draw a horizontal line from the new point to the tower to create a right-angled triangle. This forms a rectangle and a new smaller triangle, simplifying the problem.

Question 13. A guard observes an enemy boat, from an observation tower at height of 180 m above sea-level, to be at an angle of depression of 29°.
(i) Calculate to the nearest metre. the distance of the boat from the foot of the observation tower.
(ii) After some time it is observed that the boat is 200 m from the foot of the observation tower. Calculate the new angle of depression.
Answer: Let TR be the observation tower, with height TR = 180 m. Let R be the foot of the tower.
(i) Initial distance of the boat:
Let B be the initial position of the boat. The angle of depression from the top of the tower (T) to the boat (B) is \( 29^\circ \). This means the angle of elevation from the boat (B) to the top of the tower (T) is also \( 29^\circ \) (alternate interior angles).
In the right-angled triangle TRB:
\( \tan 29^\circ = \frac{\text{TR}}{\text{RB}} \)
\( \implies 0.55431 = \frac{180}{\text{RB}} \)
\( \implies \text{RB} = \frac{180}{0.55431} \approx 324.728 \text{ m} \)
Rounding to the nearest metre, the distance of the boat from the foot of the tower is 325 metres.
(ii) New angle of depression:
Let A be the new position of the boat. The new distance from the foot of the tower (R) to the boat (A) is RA = 200 m. We need to find the new angle of depression, \( \theta \), from T to A. This angle is equal to \( \angle TRA \).
In the right-angled triangle TRA:
\( \tan \theta = \frac{\text{TR}}{\text{RA}} \)
\( \implies \tan \theta = \frac{180}{200} \)
\( \implies \tan \theta = 0.9 \)
From trigonometric tables, the angle whose tangent is 0.9 is approximately \( 42^\circ \) (since \( \tan 42^\circ \approx 0.9004 \)).
So, the new angle of depression is approximately \( 42^\circ \). This shows how the angle of depression changes as the object moves closer.
In simple words: A guard on a 180-metre tower sees a boat at a 29-degree angle of depression. First, we find the boat's distance from the tower, which is 325 metres. Later, the boat moves and is now 200 metres from the tower. We then calculate the new angle of depression, which is steeper at 42 degrees.

🎯 Exam Tip: Pay close attention to alternate angles formed by the line of sight and the horizontal line. This conversion from angle of depression to angle of elevation in the triangle is crucial for solving such problems correctly.

Question 14. In the figure, the shadow of a vertical tower on level ground increases by 10 m, when the altitude of the sun changes from 45° to 30°. using the given figure, find the height or the tower, correct to one place of decimals.
Answer: Let TR be the height of the tower (let it be \( h \)). Let R be the foot of the tower. When the sun's altitude (angle of elevation) is \( 45^\circ \), the shadow is RA. When the sun's altitude changes to \( 30^\circ \), the shadow increases by 10 m, so the new shadow is RB = RA + 10. Thus, AB = 10 m.
Let RA be \( x \) metres.
First, consider the right-angled triangle TRA (when the angle is \( 45^\circ \)):
\( \tan 45^\circ = \frac{\text{TR}}{\text{RA}} \)
\( \implies 1 = \frac{h}{x} \)
\( \implies h = x \) (Equation 1)
Next, consider the right-angled triangle TRB (when the angle is \( 30^\circ \)). The base RB = RA + AB = \( x+10 \).
\( \tan 30^\circ = \frac{\text{TR}}{\text{RB}} \)
\( \implies \frac{1}{\sqrt{3}} = \frac{h}{x+10} \)
Substitute \( x = h \) from Equation 1 into this equation:
\( \implies \frac{1}{\sqrt{3}} = \frac{h}{h+10} \)
Cross-multiply:
\( \implies h+10 = h\sqrt{3} \)
Rearrange terms to solve for \( h \):
\( \implies 10 = h\sqrt{3} - h \)
\( \implies 10 = h(\sqrt{3} - 1) \)
\( \implies h = \frac{10}{\sqrt{3} - 1} \)
Using \( \sqrt{3} \approx 1.732 \):
\( \implies h = \frac{10}{1.732 - 1} = \frac{10}{0.732} \)
\( \implies h \approx 13.661 \text{ m} \)
Rounding to one decimal place, the height of the tower is 13.7 metres. This problem illustrates how shadow lengths are related to the sun's position.
In simple words: When the sun is high (45 degrees), a tower casts a certain shadow. When the sun is lower (30 degrees), the shadow becomes 10 metres longer. We use these angles to find the tower's height. After calculations, the tower is found to be 13.7 metres tall.

🎯 Exam Tip: Always pay attention to the rounding instructions (e.g., "one decimal place") at the end of the question and ensure your final answer adheres to it.

Question 15. A player sitting on the top of a tower of height 20 m observes the angle of depression of a ball lying on the ground as 60°. Find the distance between the foot of the tower and the ball. (Take \( \sqrt{3} = 1.732 \)).
Answer: Let PQ be the tower, with height PQ = 20 m. Let Q be the foot of the tower. Let B be the position of the ball on the ground. The angle of depression from the top of the tower (P) to the ball (B) is \( 60^\circ \). This means the angle of elevation from the ball (B) to the top of the tower (P) is also \( 60^\circ \) (alternate interior angles). We need to find the distance between the foot of the tower and the ball, which is BQ (let's call it \( x \)).
In the right-angled triangle PQB:
\( \tan 60^\circ = \frac{\text{PQ}}{\text{BQ}} \)
\( \implies \sqrt{3} = \frac{20}{x} \)
\( \implies x = \frac{20}{\sqrt{3}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{3}}{\sqrt{3}} \):
\( \implies x = \frac{20\sqrt{3}}{3} \)
Using \( \sqrt{3} = 1.732 \):
\( \implies x = \frac{20 \times 1.732}{3} \)
\( \implies x = \frac{34.64}{3} \)
\( \implies x \approx 11.5466 \text{ m} \)
Rounding to two decimal places (as shown in the source solution):
\( \implies x \approx 11.55 \text{ m} \)
The distance between the foot of the tower and the ball is 11.55 metres. This calculation is a simple application of trigonometry to find a horizontal distance.
In simple words: From the top of a 20-metre tower, a player sees a ball on the ground at a 60-degree downward angle. To find how far the ball is from the tower's base, we use the tangent rule and the value of square root 3 (1.732). The distance turns out to be about 11.55 metres.

🎯 Exam Tip: Always use the provided value for \( \sqrt{3} \) (if given) in your calculations to match the expected answer. Remember to rationalize denominators for cleaner expressions.

Question 16. The angle of elevation of the top of a tower from two points at distances a and b metres from the base and in the same straight line with it are complementary. Prove that the height of the tower is \( \sqrt{ab} \) metres.
Answer: Let PQ be the tower, with height PQ = \( h \). Let Q be the foot of the tower. Let A and B be two points on the same straight line from the base of the tower. The distances from the base are AQ = \( a \) metres and BQ = \( b \) metres. The angles of elevation from A and B to the top of the tower (P) are complementary. This means if \( \angle A = \theta \), then \( \angle B = 90^\circ - \theta \).
First, consider the right-angled triangle PQA (from point A):
\( \tan A = \frac{\text{PQ}}{\text{AQ}} \)
\( \implies \tan \theta = \frac{h}{a} \) (Equation 1)
Next, consider the right-angled triangle PQB (from point B):
\( \tan B = \frac{\text{PQ}}{\text{BQ}} \)
\( \implies \tan (90^\circ - \theta) = \frac{h}{b} \)
We know that \( \tan (90^\circ - \theta) = \cot \theta \).
\( \implies \cot \theta = \frac{h}{b} \) (Equation 2)
Now, multiply Equation 1 and Equation 2:
\( (\tan \theta) \times (\cot \theta) = \left( \frac{h}{a} \right) \times \left( \frac{h}{b} \right) \)
We know that \( \tan \theta \times \cot \theta = 1 \).
\( \implies 1 = \frac{h^2}{ab} \)
\( \implies h^2 = ab \)
\( \implies h = \sqrt{ab} \)
Hence, the height of the tower is \( \sqrt{ab} \) metres. This is a classic example of using complementary angles in trigonometry.
In simple words: Imagine a tower and two points on the ground, 'a' metres and 'b' metres away from its base. If the angles you look up at the tower from these points add up to 90 degrees (meaning they are complementary), then the tower's height will always be the square root of 'a' multiplied by 'b'. We proved this using basic trigonometry.

🎯 Exam Tip: When proving trigonometric identities or relationships, remember key identities like \( \tan (90^\circ - \theta) = \cot \theta \) and \( \tan \theta \times \cot \theta = 1 \). These are fundamental in many height and distance problems.

Question 17. The length of a siring between a kite and a point on the ground is 90 metres. If the string makes an angle \( \theta \) with the level ground such that \( \tan \theta = \frac{15}{8} \), how high is the kite ? Assume, there is so slack in the string.
Answer: Let K be the position of the kite and T be the point directly below the kite on the ground, so KT is the height of the kite. Let A be the point on the ground where the string is attached. So AK is the length of the string, which is 90 m. The string makes an angle \( \theta \) with the ground, and we are given \( \tan \theta = \frac{15}{8} \). We need to find the height of the kite, KT (let's call it \( h \)).
In the right-angled triangle KTA:
We know that \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\text{KT}}{\text{TA}} \). So, \( \frac{h}{\text{TA}} = \frac{15}{8} \).
This implies that for some scaling factor, if KT = 15 units, then TA = 8 units. Let's use these proportional values to find the string length (hypotenuse) in terms of these units.
If KT = 15 units and TA = 8 units, then AK (hypotenuse) \( = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \) units.
So, we have a ratio: \( \frac{\text{Height (KT)}}{\text{String Length (AK)}} = \frac{15}{17} \). This ratio is actually \( \sin \theta \).
Since the actual string length is 90 m, we can find the actual height:
\( \sin \theta = \frac{h}{90} \)
\( \implies \frac{15}{17} = \frac{h}{90} \)
\( \implies h = \frac{15 \times 90}{17} \)
\( \implies h = \frac{1350}{17} \)
\( \implies h \approx 79.411 \text{ m} \)
Rounding to one decimal place, the height of the kite is 79.4 metres. This method uses the given tangent ratio to find other trigonometric ratios needed for the problem.
In simple words: A kite string is 90 metres long. The angle it makes with the ground has a "tangent" value of 15/8. We use this ratio to find the height of the kite. By understanding that if the height is 15 parts, the base is 8 parts, then the string is 17 parts. Since the real string is 90 metres, we can figure out the real height, which is 79.4 metres.

🎯 Exam Tip: When given a ratio like \( \tan \theta = \frac{a}{b} \), you can imagine a right-angled triangle with opposite side \( a \) and adjacent side \( b \). Then use the Pythagorean theorem to find the hypotenuse, which allows you to find \( \sin \theta \) or \( \cos \theta \) as well.

Question 18. An observer in figure, \( 1\frac { 1 }{ 2 } \) m tall, is 28.5 m away from a tower 30 m high. Determine the angle of elevation from his eye to the top of the tower.
Answer: Let AB be the tower and CD be the man. The height of the tower is 30 m, and the man's height CD is \( 1\frac { 1 }{ 2 } \) m, which is 1.5 m.
The man is 28.5 m away from the foot of the tower, so CA = 28.5 m.
We draw a line DE parallel to CA. This means EA = 1.5 m and EB = 30 - 1.5 = 28.5 m.
Also, DE = CA = 28.5 m.
Now, in the right-angled triangle BED:
\( \tan \theta = \frac { BE }{ DE } = \frac { 28.5 }{ 28.5 } = 1 \)
From trigonometry tables, we know that \( \tan 45^\circ = 1 \).
So, \( \theta = 45^\circ \).
This means the angle of elevation from the observer's eye to the top of the tower is \( 45^\circ \). A \( 45^\circ \) angle means the horizontal distance equals the vertical height difference.
In simple words: The man is 1.5 meters tall and stands 28.5 meters from a 30-meter tower. We find the height from his eye to the tower's top is 28.5 meters, and the distance to the tower is also 28.5 meters. Since these two lengths are equal, the angle to the top is 45 degrees.

🎯 Exam Tip: Always draw a clear diagram and label all given measurements. Remember to subtract the observer's height from the total height of the object to find the relevant vertical distance for the angle of elevation calculation.

Question 19. Two men are on diametrically opposite side of a tower. They measure the angles of elevation of the top of the tower as 20° and 24° respectively. If the height of the tower is 40 m, find the distance between them.
Answer: Let TR be the tower with a height of 40 m. Let A and B be the positions of the two men on opposite sides of the tower.
The angle of elevation from point A to the top of the tower is \( 20^\circ \). Let AR = x.
The angle of elevation from point B to the top of the tower is \( 24^\circ \). Let BR = y.
In the right-angled triangle TRA:
\( \tan 20^\circ = \frac { TR }{ AR } \)
\( 0.36397 = \frac { 40 }{ x } \)
\( x = \frac { 40 }{ 0.36397 } = 109.89 \) m
In the right-angled triangle TRB:
\( \tan 24^\circ = \frac { TR }{ BR } \)
\( 0.44523 = \frac { 40 }{ y } \)
\( y = \frac { 40 }{ 0.44523 } = 89.84 \) m
The total distance between the two men is \( x + y \).
Distance \( = 109.89 + 89.84 = 199.73 \) m.
This problem uses basic trigonometry to find unknown distances from known angles and heights. It's a practical application of angles of elevation.
In simple words: A tower is 40 meters high. Two people stand on opposite sides and look up at it. One person sees it at a 20-degree angle, and the other at a 24-degree angle. We use these angles to find how far each person is from the tower, and then add those distances to find how far apart they are from each other, which is 199.73 meters.

🎯 Exam Tip: When two observers are on opposite sides, their distances from the base of the tower are added to find the total distance between them. Always use the correct trigonometric ratio for the given angle and sides.

Question 20. The horizontal distance between the two towers is 60 m and the angular depression of the top of the first, as seen from the top of the second, which is 150 m high, is 30°; find the height of the first.
Answer: Let AB be the first tower (height h) and CD be the second tower (height 150 m). The horizontal distance between them, BD, is 60 m.
We draw a line AE parallel to BD. This means AE = BD = 60 m.
Also, ED = AB = h. Therefore, CE = CD - ED = 150 - h.
The angle of depression from the top of the second tower (C) to the top of the first tower (A) is \( 30^\circ \).
Because AE is parallel to BD and CX is the horizontal line through C, the alternate angle \( \angle CAE \) is equal to the angle of depression, so \( \angle CAE = 30^\circ \).
Now, in the right-angled triangle CAE:
\( \tan 30^\circ = \frac { CE }{ AE } \)
\( \frac{1}{\sqrt{3}} = \frac { 150-h }{ 60 } \)
\( \sqrt{3}(150-h) = 60 \)
\( 150-h = \frac{60}{\sqrt{3}} \)
\( 150-h = \frac{60 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \)
\( 150-h = \frac{60 \sqrt{3}}{3} = 20\sqrt{3} \)
Using \( \sqrt{3} = 1.732 \):
\( 150-h = 20 \times 1.732 = 34.64 \)
\( h = 150 - 34.64 = 115.36 \) m
The height of the first tower AB is 115.36 m. Understanding alternate angles is key to solving such problems efficiently.
In simple words: There are two towers 60 meters apart. The second tower is 150 meters tall. If you look down from the top of the second tower to the top of the first tower, the angle is 30 degrees. We use this to find the height of the first tower, which comes out to be 115.36 meters.

🎯 Exam Tip: When dealing with angles of depression between two objects of different heights, always draw a horizontal line from the higher point. This creates a right-angled triangle and helps identify the correct angles (using alternate interior angles).

Question 21. From the top of a cliff, 150 metres high, the angle of depression of the top and bottom of a tower are observed to be 30° and 60°, find the height of the tower.
Answer: Let AB be the cliff, which is 150 m high. Let CD be the tower (height h).
The angle of depression from the top of the cliff (A) to the top of the tower (C) is \( 30^\circ \).
The angle of depression from the top of the cliff (A) to the bottom of the tower (D) is \( 60^\circ \).
Let's draw a horizontal line AX from A. Since AX is parallel to BD, by alternate interior angles:
\( \angle XAC = \angle ACE = 30^\circ \)
\( \angle XAD = \angle ADB = 60^\circ \)
In the right-angled triangle ABD:
\( \tan 60^\circ = \frac { AB }{ BD } \)
\( \sqrt{3} = \frac { 150 }{ BD } \)
\( BD = \frac { 150 }{ \sqrt{3} } = \frac { 150 \sqrt{3} }{ 3 } = 50\sqrt{3} \) m
Now, draw a horizontal line CE from C to AB. Then CE = BD = \( 50\sqrt{3} \) m.
Also, EB = CD = h. So, AE = AB - EB = 150 - h.
In the right-angled triangle ACE:
\( \tan 30^\circ = \frac { AE }{ CE } \)
\( \frac{1}{\sqrt{3}} = \frac { 150-h }{ 50\sqrt{3} } \)
\( 50\sqrt{3} = \sqrt{3}(150-h) \)
Divide by \( \sqrt{3} \) on both sides:
\( 50 = 150 - h \)
\( h = 150 - 50 = 100 \) m
The height of the tower is 100 m. This method is often used to determine unknown heights or distances in surveying.
In simple words: From a 150-meter-high cliff, a person sees the top of a tower with a 30-degree downward angle and the bottom of the tower with a 60-degree downward angle. We use these angles to calculate that the tower is 100 meters tall.

🎯 Exam Tip: For problems involving angles of depression from a high point to both the top and bottom of a tower, always define the horizontal distance between the cliff and the tower first. This forms two right-angled triangles that share a common side (the horizontal distance).

Question 22. From a point P on level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, how far is P from the foot of the tower?
Answer: Let AB be the tower with a height of 100 m. Let P be a point on the level ground.
The angle of elevation from point P to the top of the tower (A) is \( 30^\circ \).
We need to find the distance PB, which we'll call x.
In the right-angled triangle ABP:
\( \tan 30^\circ = \frac { AB }{ PB } \)
\( \frac{1}{\sqrt{3}} = \frac { 100 }{ x } \)
\( x = 100\sqrt{3} \)
Using \( \sqrt{3} = 1.732 \):
\( x = 100 \times 1.732 = 173.2 \) m
So, the distance from point P to the foot of the tower is 173.2 m. This calculation is common in construction and engineering.
In simple words: A tower is 100 meters tall. From a spot on the ground, someone looks up at the top of the tower, and the angle is 30 degrees. To find out how far that spot is from the base of the tower, we use simple math and find it is 173.2 meters away.

🎯 Exam Tip: This is a straightforward application of the tangent ratio. Always draw and label the right-angled triangle correctly to avoid errors. Remember that the tangent of an angle is the ratio of the opposite side to the adjacent side.

Question 23. From the top of a building AB, 60 metres high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find
(i) the horizontal distance between AB and CD;
(ii) the height of the lamp post CD.
Answer: Let AB be the building (height 60 m) and CD be the lamp post (height h).
From the top of the building (A), the angle of depression to the top of the lamp post (C) is \( 30^\circ \).
The angle of depression to the bottom of the lamp post (D) is \( 60^\circ \).
(i) To find the horizontal distance between AB and CD:
Draw a line AE parallel to the ground (BD). Then AE = BD (horizontal distance).
From alternate interior angles:
\( \angle ADB = 60^\circ \)
In the right-angled triangle ABD:
\( \tan 60^\circ = \frac { AB }{ BD } \)
\( \sqrt{3} = \frac { 60 }{ BD } \)
\( BD = \frac { 60 }{ \sqrt{3} } = \frac { 60\sqrt{3} }{ 3 } = 20\sqrt{3} \) m
Using \( \sqrt{3} = 1.732 \):
\( BD = 20 \times 1.732 = 34.64 \) m
The horizontal distance between the building and the lamp post is 34.64 m.
(ii) To find the height of the lamp post CD:
We know CE = BD = \( 20\sqrt{3} \) m.
Also, EB = CD = h. So, AE = AB - EB = 60 - h.
From alternate interior angles, \( \angle ACE = 30^\circ \).
In the right-angled triangle ACE:
\( \tan 30^\circ = \frac { AE }{ CE } \)
\( \frac{1}{\sqrt{3}} = \frac { 60-h }{ 20\sqrt{3} } \)
\( 20\sqrt{3} = \sqrt{3}(60-h) \)
Divide by \( \sqrt{3} \) on both sides:
\( 20 = 60 - h \)
\( h = 60 - 20 = 40 \) m
The height of the lamp post CD is 40 m. This illustrates how angles of depression can be used to find heights and distances.
In simple words: From the top of a 60-meter building, a person looks down at a lamp post. The top of the post is seen at a 30-degree angle, and the bottom at a 60-degree angle. First, we find the horizontal distance between the building and the post is 34.64 meters. Then, we find the lamp post is 40 meters tall.

🎯 Exam Tip: When given angles of depression to the top and bottom of an object, always draw a horizontal line from the observer's eye. This creates two right-angled triangles, one for the angle to the top and one for the angle to the bottom, sharing the horizontal distance as a common side.

Question 24. An aeroplane when 34000 in high passes vertically above another aeroplane at an instance when their angles of elevation at the same observation pout are 60° and 45°. How many metres higher is the one than the other?
Answer: Let D be the observation point on the ground. Let A be the higher aeroplane and C be the lower aeroplane, both vertically above point B on the ground.
The height of the higher aeroplane (AD) is 3000 m (note: the question states 34000 but the solution uses 3000, we will follow the solution's consistent use of 3000 m).
The angle of elevation of A from D is \( 60^\circ \).
The angle of elevation of C from D is \( 45^\circ \).
Let DB = x m and let the distance between the two aeroplanes be h m. Then CD = AD - h = 3000 - h.
In the right-angled triangle ABD:
\( \tan 60^\circ = \frac { AD }{ DB } \)
\( \sqrt{3} = \frac { 3000 }{ x } \)
\( x = \frac { 3000 }{ \sqrt{3} } = \frac { 3000\sqrt{3} }{ 3 } = 1000\sqrt{3} \) m
In the right-angled triangle CBD:
\( \tan 45^\circ = \frac { CD }{ DB } \)
\( 1 = \frac { 3000-h }{ x } \)
So, \( x = 3000 - h \)
Now, we can substitute the value of x:
\( 1000\sqrt{3} = 3000 - h \)
\( h = 3000 - 1000\sqrt{3} \)
Using \( \sqrt{3} = 1.732 \):
\( h = 3000 - (1000 \times 1.732) \)
\( h = 3000 - 1732 = 1268 \) m
The first aeroplane is 1268 m higher than the second. This demonstrates how trigonometric ratios help in determining relative heights.
In simple words: An aeroplane is 3000 meters high. Another aeroplane is directly below it. From the ground, someone sees the top plane at a 60-degree angle and the bottom plane at a 45-degree angle. We use these angles to find that the top plane is 1268 meters higher than the bottom one.

🎯 Exam Tip: When dealing with objects directly above each other, set up two right-angled triangles sharing the same horizontal base. Calculate the base distance using the larger angle (and greater height) first, then use it to find the unknown height of the lower object or the difference in heights.

Question 25. The angle of elevation of a cloud from a point 60 m above the lake is 30° and the angle of depression of the reflection of the cloud in the lake is 60°. Find the height of the cloud.
Answer: Let P be the observation point, 60 m above the lake surface RS. Let C be the cloud and T be its reflection in the lake.
Draw PQ perpendicular to CS. Then PQ is the horizontal distance from P to the vertical line passing through the cloud.
Let CQ = h. So, the total height of the cloud from the lake surface is CS = CQ + QS = h + 60 m.
The reflection T will be at the same depth below the lake surface as the cloud is above it. So, ST = CS = h + 60 m.
From point P, the angle of elevation to the cloud C is \( 30^\circ \). In \( \triangle CPQ \):
\( \tan 30^\circ = \frac { CQ }{ PQ } \)
\( \frac{1}{\sqrt{3}} = \frac { h }{ PQ } \)
\( PQ = h\sqrt{3} \) ----(i)
From point P, the angle of depression to the reflection T is \( 60^\circ \). The vertical distance from P to T is QT = QS + ST = 60 + (h + 60) = h + 120 m.
In \( \triangle TPQ \):
\( \tan 60^\circ = \frac { QT }{ PQ } \)
\( \sqrt{3} = \frac { h+120 }{ PQ } \)
\( PQ = \frac { h+120 }{ \sqrt{3} } \) ----(ii)
Equating (i) and (ii):
\( h\sqrt{3} = \frac { h+120 }{ \sqrt{3} } \)
Multiply both sides by \( \sqrt{3} \):
\( 3h = h + 120 \)
\( 2h = 120 \)
\( h = 60 \) m
The height of the cloud from the lake surface is CS = h + 60 = 60 + 60 = 120 m. This problem applies the principle that the angle of incidence equals the angle of reflection.
In simple words: From 60 meters above a lake, a cloud is seen at a 30-degree angle up, and its reflection in the lake is seen at a 60-degree angle down. We use these angles to find that the cloud is 120 meters above the lake.

🎯 Exam Tip: For problems involving reflections in water, remember that the depth of the reflection below the surface is equal to the height of the object above the surface. This creates a symmetrical setup for calculations.

Question 26. The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of elevation of the top of the tower fro the foot of the hill is 30°. If the tower 50 m high, what is the height of the hill?
Answer: Let AB be the hill with height h, and CD be the tower with height 50 m.
The angle of elevation of the top of the hill (A) from the foot of the tower (D) is \( 60^\circ \).
The angle of elevation of the top of the tower (C) from the foot of the hill (B) is \( 30^\circ \).
Let the horizontal distance between the hill and the tower, BD, be x m.
In the right-angled triangle CDB:
\( \tan 30^\circ = \frac { CD }{ BD } \)
\( \frac{1}{\sqrt{3}} = \frac { 50 }{ x } \)
\( x = 50\sqrt{3} \) m ----(i)
In the right-angled triangle ABD:
\( \tan 60^\circ = \frac { AB }{ BD } \)
\( \sqrt{3} = \frac { h }{ x } \)
\( h = x\sqrt{3} \) ----(ii)
Substitute the value of x from (i) into (ii):
\( h = (50\sqrt{3})\sqrt{3} \)
\( h = 50 \times 3 \)
\( h = 150 \) m
The height of the hill is 150 m. This solution elegantly connects two different angles of elevation through a common base.
In simple words: From the bottom of a 50-meter tower, the top of a hill is seen at a 60-degree angle up. From the bottom of that hill, the top of the tower is seen at a 30-degree angle up. Using these facts, we figure out that the hill is 150 meters tall.

🎯 Exam Tip: When angles of elevation are given from the base of one object to the top of another, and vice-versa, always establish a common horizontal base. This creates two interconnected right-angled triangles that can be solved simultaneously.

Self Evaluation And Revision (Latest ICSE Questions)

Question 1. From a window A, 10 m above ground, the angle of elevation of the top C of a tower is x°, where tan x = \( \frac { 5 }{ 2 } \), and the angle of depression of the foot D of the tower is y°, where tan y= \( \frac { 1 }{ 4 } \). See figure given here. Calculate the height CD of the tower in metres.
Answer: Let CD be the tower and A be the window 10 m above the ground BD.
The height of the window from the ground, AB, is 10 m.
Draw AE parallel to BD. Then ED = AB = 10 m and AE = BD.
Let CD = h. Then CE = CD - ED = h - 10.
Given, the angle of elevation of the top C of the tower from A is x°, and \( \tan x = \frac { 5 }{ 2 } \).
In the right-angled triangle ACE:
\( \tan x = \frac { CE }{ AE } \)
\( \frac { 5 }{ 2 } = \frac { h-10 }{ AE } \)
\( AE = \frac { 2(h-10) }{ 5 } \)
Given, the angle of depression of the foot D of the tower from A is y°, and \( \tan y = \frac { 1 }{ 4 } \).
In the right-angled triangle ABD:
\( \tan y = \frac { AB }{ BD } \)
\( \frac { 1 }{ 4 } = \frac { 10 }{ BD } \)
\( BD = 40 \) m
Since AE = BD, we have \( AE = 40 \) m.
Now substitute AE = 40 into the equation for AE:
\( 40 = \frac { 2(h-10) }{ 5 } \)
\( 40 \times 5 = 2(h-10) \)
\( 200 = 2h - 20 \)
\( 2h = 200 + 20 \)
\( 2h = 220 \)
\( h = 110 \) m
The height of the tower CD is 110 m. This problem combines both angles of elevation and depression effectively.
In simple words: From a window 10 meters high, the top of a tower is seen with a certain upward angle (tan x = 5/2), and the bottom of the tower is seen with a certain downward angle (tan y = 1/4). We use these angles and the window's height to find that the tower is 110 meters tall.

🎯 Exam Tip: When given specific tangent values for angles, use them directly in the trigonometric ratios rather than finding the angles themselves. This avoids rounding errors that could occur if you calculate the angle first.

Question 2. In the figure, \( \angle PSR = 90^\circ \), PQ = 10 cm, QS = 6 cm, RQ = 9 cm. Calculate the length of PR.
Answer: Given the figure, we have:
\( \angle PSR = 90^\circ \)
PQ = 10 cm
QS = 6 cm
RQ = 9 cm
First, find the length of RS:
RS = RQ + QS = 9 cm + 6 cm = 15 cm.
Now, in the right-angled triangle PQS, we can find PS using the Pythagorean theorem:
\( PS^2 + QS^2 = PQ^2 \)
\( PS^2 + 6^2 = 10^2 \)
\( PS^2 + 36 = 100 \)
\( PS^2 = 100 - 36 \)
\( PS^2 = 64 \)
\( PS = \sqrt{64} = 8 \) cm.
Now, consider the right-angled triangle PRS:
\( PR^2 = PS^2 + RS^2 \)
\( PR^2 = 8^2 + 15^2 \)
\( PR^2 = 64 + 225 \)
\( PR^2 = 289 \)
\( PR = \sqrt{289} = 17 \) cm.
The length of PR is 17 cm. The Pythagorean theorem is a fundamental tool in geometry, especially for right triangles.
In simple words: We have a shape with some lengths and a right angle. We first find the total length of one side by adding two given parts (RS = 9 + 6 = 15 cm). Then, we use the Pythagorean theorem on a smaller triangle to find a missing side (PS = 8 cm). Finally, we use the Pythagorean theorem again on the larger triangle to find the length of PR, which is 17 cm.

🎯 Exam Tip: Break down complex figures into simpler right-angled triangles. Always identify the hypotenuse correctly for the Pythagorean theorem. Double-check your arithmetic, especially when squaring numbers.

Question 3. A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower?
Answer: Let BC be the vertical tower, which is 20 m high. Let A be the man's position on the ground.
The angle of elevation to the top of the tower (C) from the man's position (A) is \( \theta \).
We are given that \( \cos \theta = 0.53 \).
In the right-angled triangle ABC:
\( \cos \theta = \frac { AB }{ AC } \)
We need to find the distance AB, which is how far the man is from the foot of the tower.
Using a calculator or trigonometric tables, if \( \cos \theta = 0.53 \), then \( \theta \approx 58^\circ \).
Now we can use the tangent ratio:
\( \tan \theta = \frac { BC }{ AB } \)
\( \tan 58^\circ = \frac { 20 }{ AB } \)
From tables, \( \tan 58^\circ \approx 1.6000 \).
\( 1.6000 = \frac { 20 }{ AB } \)
\( AB = \frac { 20 }{ 1.6000 } = \frac { 200 }{ 16 } = 12.5 \) m
So, the man is standing 12.5 m away from the foot of the tower. Cosine and tangent are related functions, and using either can lead to the correct distance.
In simple words: A tower is 20 meters high. A man looks at the top of the tower. He knows that the cosine of the angle he is looking up at is 0.53. From this, we first find the angle is about 58 degrees. Then, using the tangent of that angle, we calculate that he is standing 12.5 meters away from the bottom of the tower.

🎯 Exam Tip: Be comfortable converting between trigonometric ratios or finding angles from given ratios. If you are given a cosine or sine value and need to use tangent, it's often easiest to find the angle first, or construct a right-angled triangle to find the third side.

Question 4. The shadow of a vertical tower AB on level ground is increased by 10 m, when the altitude changes from 45° to 30°, as shown in the adjoining figure. Find the height of the tower and give your answer correct to \( \frac { 1 }{ 10 } \) of a metre.
Answer: Let AB be the vertical tower with height h.
Let CB be the length of its shadow when the sun's altitude (angle of elevation) is \( 45^\circ \).
Let DB be the length of its shadow when the sun's altitude is \( 30^\circ \).
Given that the shadow length increases by 10 m, so DC = 10 m.
This means DB = CB + 10.
In the right-angled triangle ACB:
\( \tan 45^\circ = \frac { AB }{ CB } \)
\( 1 = \frac { h }{ CB } \)
So, \( CB = h \) ----(i)
In the right-angled triangle ADB:
\( \tan 30^\circ = \frac { AB }{ DB } \)
\( \frac{1}{\sqrt{3}} = \frac { h }{ DB } \)
Substitute DB = CB + 10 into this equation:
\( \frac{1}{\sqrt{3}} = \frac { h }{ h+10 } \) (using \( CB=h \))
Cross-multiply:
\( h+10 = h\sqrt{3} \)
Rearrange to solve for h:
\( 10 = h\sqrt{3} - h \)
\( 10 = h(\sqrt{3} - 1) \)
\( h = \frac { 10 }{ \sqrt{3} - 1 } \)
Using \( \sqrt{3} = 1.732 \):
\( h = \frac { 10 }{ 1.732 - 1 } \)
\( h = \frac { 10 }{ 0.732 } \)
\( h = 13.661... \)
Rounding to one decimal place (nearest \( \frac { 1 }{ 10 } \) of a metre):
\( h = 13.7 \) m.
The height of the tower is 13.7 m. Such problems highlight how shadow lengths change with the sun's position.
In simple words: When the sun is high (45 degrees), a tower's shadow is shorter. When the sun is lower (30 degrees), its shadow gets 10 meters longer. We use these angles and the change in shadow length to calculate the tower's height, which is 13.7 meters.

🎯 Exam Tip: Always make sure to rationalise the denominator if your intermediate answer contains a square root. For "correct to \( \frac{1}{10} \) of a metre", it means rounding to one decimal place.

Question 5. A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, he finds the angle of elevation to be 30°. Calculate:
(i) the width of the river and
(ii) the height of the tree.
Answer: Let TR be the tree (height h) on one bank, and R be its base. Let A be the initial position of the man on the opposite bank and B be his position after moving 50 m away from the bank.
The width of the river is AR = x m.
The man moves 50 m away, so AB = 50 m. The new distance from the tree is BR = AR + AB = x + 50 m.
(i) To calculate the width of the river (x):
From the initial position A, the angle of elevation to the top of the tree (T) is \( 60^\circ \). In \( \triangle TRA \):
\( \tan 60^\circ = \frac { TR }{ AR } \)
\( \sqrt{3} = \frac { h }{ x } \)
\( h = x\sqrt{3} \) ----(i)
From the new position B, the angle of elevation to the top of the tree (T) is \( 30^\circ \). In \( \triangle TRB \):
\( \tan 30^\circ = \frac { TR }{ BR } \)
\( \frac{1}{\sqrt{3}} = \frac { h }{ x+50 } \)
\( h\sqrt{3} = x+50 \) ----(ii)
Substitute \( h = x\sqrt{3} \) from (i) into (ii):
\( (x\sqrt{3})\sqrt{3} = x+50 \)
\( 3x = x+50 \)
\( 2x = 50 \)
\( x = 25 \) m
The width of the river is 25 m.
(ii) To calculate the height of the tree (h):
Substitute x = 25 m back into equation (i):
\( h = x\sqrt{3} = 25\sqrt{3} \)
Using \( \sqrt{3} = 1.732 \):
\( h = 25 \times 1.732 = 43.3 \) m
The height of the tree is 43.3 m. This common type of problem demonstrates using two observations to find unknown dimensions.
In simple words: A man sees a tree across a river at a 60-degree angle. He then walks 50 meters back and sees the tree at a 30-degree angle. We use these angles to find two things: first, the river is 25 meters wide, and second, the tree is 43.3 meters tall.

🎯 Exam Tip: Always set up two equations from the two observation points and solve them simultaneously. It's often helpful to express the unknown height (h) in terms of the unknown distance (x) from one triangle, then substitute it into the second equation.

Question 6. Two people standing on the same side of a tower in straight line with it, measure the angles of elevation of the top of the tower as 25° and 50° respectively. If the height of the tower is 70°, find the distance between the two people.
Answer: Let TR be the tower with height 70 m. Let A and B be the positions of the two people, standing on the same side of the tower.
The person closer to the tower (B) measures an angle of elevation of \( 50^\circ \) to the top of the tower (T). Let BR = x m.
The person farther from the tower (A) measures an angle of elevation of \( 25^\circ \) to the top of the tower (T). Let AR = y m.
We need to find the distance between the two people, which is AB = AR - BR = y - x.
In the right-angled triangle TRB:
\( \tan 50^\circ = \frac { TR }{ BR } \)
\( 1.19175 = \frac { 70 }{ x } \)
\( x = \frac { 70 }{ 1.19175 } = 58.737 \) m
In the right-angled triangle TRA:
\( \tan 25^\circ = \frac { TR }{ AR } \)
\( 0.46631 = \frac { 70 }{ y } \)
\( y = \frac { 70 }{ 0.46631 } = 150.115 \) m
The distance between the two people is \( AB = y - x \):
\( AB = 150.115 - 58.737 = 91.378 \) m
Rounding to two decimal places, the distance is 91.38 m. This illustrates how angles of elevation from different points can help calculate distances.
In simple words: A 70-meter tower is observed by two people standing on the same side. The person closer to the tower sees its top at a 50-degree angle, and the one farther away sees it at a 25-degree angle. We calculate each person's distance from the tower and then subtract to find they are 91.38 meters apart.

🎯 Exam Tip: When observers are on the same side of an object, you find the difference in their distances from the base of the object. Make sure to use accurate trigonometric values for non-standard angles (e.g., from tables or a calculator) and keep precision during calculations before final rounding.

Question 7. From the top of a cliff 92 m high, the angle of depression of a boy is 20°. Calculate to the nearest metre, the distance of the boy from the foot of the cliff.
Answer: Let AB be the cliff, with a height of 92 metres. Let the boy be at point C on the ground. The angle of depression from the top of the cliff (A) to the boy (C) is 20°. This means the angle of elevation from the boy (C) to the top of the cliff (A), which is \( \angle ACB \), is also 20° because they are alternate angles. We need to find the horizontal distance BC. In the right-angled triangle \( \triangle ABC \):
\( \tan 20° = \frac{AB}{BC} \)
\( 0.36397 = \frac{92}{BC} \)
\( BC = \frac{92}{0.36397} \)
\( BC = 252.768 \) m
Rounded to the nearest metre, the distance is 253 m. Alternate interior angles are equal when two parallel lines (the horizontal line from the cliff top and the ground) are crossed by a transversal line.
In simple words: The cliff is 92 metres high. A boy on the ground looks up, and the angle from his spot to the top of the cliff is 20 degrees. We use trigonometry to find how far the boy is from the bottom of the cliff. That distance is about 253 metres.

🎯 Exam Tip: Always make a clear diagram for problems involving angles of elevation or depression. Remember that the angle of depression from point A to point C is equal to the angle of elevation from point C to point A (alternate interior angles).

Question 8. The shadow of a vertical tower on a level ground increases by 10 m, when the altitude changes from 45° to 30°. Find the height of the tower, correct to two decimal places.
Answer: Let AB be the vertical tower, and let its height be \( h \) metres. Let C be the point where the shadow ends when the sun's elevation (altitude) is 45°, and D be the point where the shadow ends when the sun's elevation is 30°. The shadow increases by 10 m, so CD = 10 m.
In the right-angled triangle \( \triangle ABC \):
\( \tan 45° = \frac{AB}{BC} \)
\( 1 = \frac{h}{BC} \)
\( BC = h \) metres (Equation 1)
In the right-angled triangle \( \triangle ABD \):
\( \tan 30° = \frac{AB}{BD} \)
\( \frac{1}{\sqrt{3}} = \frac{h}{BC + CD} \)
\( \frac{1}{\sqrt{3}} = \frac{h}{h + 10} \) (Substituting BC = h from Equation 1)
\( h + 10 = h\sqrt{3} \)
\( 10 = h\sqrt{3} - h \)
\( 10 = h(\sqrt{3} - 1) \)
Using \( \sqrt{3} \approx 1.732 \):
\( 10 = h(1.732 - 1) \)
\( 10 = h(0.732) \)
\( h = \frac{10}{0.732} \)
\( h = 13.661 \)
Rounding to two decimal places, the height of the tower is 13.66 m. A lower sun angle makes shadows longer, which is why the 30° elevation has a longer shadow.
In simple words: When the sun is high (45 degrees), the tower's shadow is short. When the sun is lower (30 degrees), the shadow becomes 10 metres longer. By using basic angle rules, we find the tower is about 13.66 metres tall.

🎯 Exam Tip: When the shadow length changes with the sun's altitude, always set up two trigonometric equations for the two different right-angled triangles formed. Solve them simultaneously to find the unknown height or distance.

Question 9. From the top of a hill the angles of depression of two consecutive kilometre stones, due east are found to be 30° and 45° respectively. Find the distance of the two stones from the foot of the hill.
Answer: Let PQ be the height of the hill, \( h \) km. Let A and B be the positions of the two consecutive kilometre stones, so the distance AB = 1 km. Since they are "due east," they are in a straight line from the foot of the hill. The angles of depression from the top of the hill (P) to the stones A and B are 30° and 45° respectively. This means the angles of elevation from the stones to the top of the hill are \( \angle PAQ = 30° \) and \( \angle PBQ = 45° \). Let the distance from the foot of the hill (Q) to the nearer stone (B) be \( x \) km. Then the distance from Q to A is \( (x+1) \) km.
In the right-angled triangle \( \triangle PBQ \):
\( \tan 45° = \frac{PQ}{BQ} \)
\( 1 = \frac{h}{x} \)
\( h = x \) (Equation 1)
In the right-angled triangle \( \triangle PAQ \):
\( \tan 30° = \frac{PQ}{AQ} \)
\( \frac{1}{\sqrt{3}} = \frac{h}{x+1} \)
\( x+1 = h\sqrt{3} \) (Equation 2)
Substitute \( h=x \) from Equation 1 into Equation 2:
\( x+1 = x\sqrt{3} \)
\( 1 = x\sqrt{3} - x \)
\( 1 = x(\sqrt{3} - 1) \)
\( x = \frac{1}{\sqrt{3} - 1} \)
To rationalize, multiply by \( \frac{\sqrt{3}+1}{\sqrt{3}+1} \):
\( x = \frac{\sqrt{3}+1}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{\sqrt{3}+1}{3-1} = \frac{\sqrt{3}+1}{2} \)
Using \( \sqrt{3} \approx 1.732 \):
\( x = \frac{1.732+1}{2} = \frac{2.732}{2} = 1.366 \) km
So, the distance of the nearer stone (B) from the foot of the hill (Q) is 1.366 km. The distance of the further stone (A) from the foot of the hill (Q) is \( x+1 = 1.366+1 = 2.366 \) km. Consecutive kilometre stones are exactly 1 km apart.
In simple words: From the top of a hill, we look down at two stones 1 km apart. The closer stone is seen at a 45-degree angle, and the further one at 30 degrees. The closer stone is 1.366 km away from the bottom of the hill, and the further stone is 2.366 km away.

🎯 Exam Tip: When dealing with objects in a straight line, carefully define your variables for the distances from the base. Using 'x' and 'x+1' (or 'x-1') helps simplify the setup of the two trigonometric equations.

Question 10. A vertical pole and a vertical tower are on the same level ground. From the top of the pole, the angle of elevation of the top of the tower is 60° and the angle of depression of the foot of the tower is 30°. Find the height of the tower if the length of pole is 20 m.
Answer: Let AE be the tower and BC be the pole. The height of the pole (BC) is 20 m. Let the height of the tower (AE) be \( h \) metres. Draw a horizontal line BD from the top of the pole (B) to meet the tower at D. This creates a rectangle CDEB, so CD = BC = 20 m and BD = CE. Let CE be \( x \) metres.
The angle of depression of the foot of the tower (E) from the top of the pole (B) is 30°. So, in the right-angled triangle \( \triangle BCE \):
\( \tan 30° = \frac{BC}{CE} \)
\( \frac{1}{\sqrt{3}} = \frac{20}{x} \)
\( x = 20\sqrt{3} \) m
So, the horizontal distance CE (and also BD) is \( 20\sqrt{3} \) m. Now, the height AD = AE - DE = AE - BC = \( (h - 20) \) m. The angle of elevation of the top of the tower (A) from the top of the pole (B) is 60°. So, in the right-angled triangle \( \triangle ABD \):
\( \tan 60° = \frac{AD}{BD} \)
\( \sqrt{3} = \frac{h-20}{20\sqrt{3}} \)
\( 20\sqrt{3} \times \sqrt{3} = h - 20 \)
\( 20 \times 3 = h - 20 \)
\( 60 = h - 20 \)
\( h = 60 + 20 \)
\( h = 80 \) m
The height of the tower is 80 m. Drawing a horizontal line from the observation point helps split the problem into easier right-angled triangles.
In simple words: Imagine a 20-metre pole and a taller tower. From the top of the pole, looking up to the tower's top is 60 degrees, and looking down to its base is 30 degrees. We use these angles to find out that the tower is 80 metres tall.

🎯 Exam Tip: For problems involving observations between two vertical objects, draw a horizontal line from the top of the lower object to create a right-angled triangle with the upper part of the taller object. This method helps simplify the geometry and calculations.

Question 11. From two points A and B on the same side of a building, the angles of elevation of the top of the building are 30° and 60° respectively. If the height of the building is 10 m, find the distance between A and B correct to two decimal places.
Answer: Let DC be the building with height 10 m. Let B be the point closer to the building, and A be the point further away, both on the same side. The angle of elevation from B to the top of the building (D) is 60° (\( \angle DBC = 60° \)), and from A to D is 30° (\( \angle DAC = 30° \)). We need to find the distance AB.
In the right-angled triangle \( \triangle DBC \):
\( \tan 60° = \frac{DC}{BC} \)
\( \sqrt{3} = \frac{10}{BC} \)
\( BC = \frac{10}{\sqrt{3}} \) metres (Equation 1)
In the right-angled triangle \( \triangle DAC \):
\( \tan 30° = \frac{DC}{AC} \)
\( \frac{1}{\sqrt{3}} = \frac{10}{AC} \)
\( AC = 10\sqrt{3} \) metres (Equation 2)
The distance between points A and B is AB = AC - BC. Subtract Equation 1 from Equation 2:
\( AB = 10\sqrt{3} - \frac{10}{\sqrt{3}} \)
To simplify, find a common denominator:
\( AB = \frac{10\sqrt{3} \times \sqrt{3} - 10}{\sqrt{3}} \)
\( AB = \frac{10 \times 3 - 10}{\sqrt{3}} \)
\( AB = \frac{30 - 10}{\sqrt{3}} = \frac{20}{\sqrt{3}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{3}}{\sqrt{3}} \):
\( AB = \frac{20\sqrt{3}}{3} \)
Using \( \sqrt{3} \approx 1.732 \):
\( AB = \frac{20 \times 1.732}{3} = \frac{34.64}{3} \)
\( AB = 11.5466... \)
Rounding to two decimal places, the distance between A and B is 11.55 m. Rationalizing the denominator gives a cleaner numerical result.
In simple words: A building is 10 metres high. Someone observes its top from two spots on the same side. From the closer spot, the angle is 60 degrees. From the further spot, it's 30 degrees. We calculate that the two spots are about 11.55 metres apart.

🎯 Exam Tip: When points are on the same side of a vertical object, the distance between them is the difference between their distances from the base. Remember to rationalize the denominator if \( \sqrt{3} \) or other radicals are in the final fraction for exact answers.

Question 12. The top of a light house 100 m high the angle of depression of two ships on opposite sides of it are 48° and 36° respectively. Find the distance between the two ships to the nearest meter.
Answer: Let AD be the lighthouse, with a height of 100 m. Let B and C be the positions of the two ships on opposite sides of the lighthouse's base (D). The angle of depression from the top of the lighthouse (A) to ship B is 48°, so \( \angle ABD = 48° \) (alternate angle). The angle of depression from A to ship C is 36°, so \( \angle ACD = 36° \) (alternate angle). We need to find the distance BC, which is BD + DC.
In the right-angled triangle \( \triangle ABD \):
\( \tan 48° = \frac{AD}{BD} \)
\( BD = \frac{AD}{\tan 48°} \)
\( BD = \frac{100}{1.1106} \) (Using \( \tan 48° \approx 1.1106 \))
\( BD = 90.041 \) m
In the right-angled triangle \( \triangle ADC \):
\( \tan 36° = \frac{AD}{DC} \)
\( DC = \frac{AD}{\tan 36°} \)
\( DC = \frac{100}{0.7265} \) (Using \( \tan 36° \approx 0.7265 \))
\( DC = 137.646 \) m
The total distance between the two ships is BC = BD + DC.
\( BC = 90.041 + 137.646 = 227.687 \) m
Rounding to the nearest metre, the distance between the ships is 228 m. The alternate angles of depression are useful for converting to angles within the triangles.
In simple words: A lighthouse is 100 metres tall. Two ships are on opposite sides of it. From the top of the lighthouse, one ship looks 48 degrees down, and the other looks 36 degrees down. We calculate how far each ship is from the lighthouse, then add those distances. The ships are about 228 metres apart.

🎯 Exam Tip: When objects are on opposite sides, the total distance is the sum of their individual distances from the base. Remember to use accurate trigonometric values to avoid rounding errors in intermediate steps.

Question 13. A man observes the angles of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m the angle of elevation changes to 60°. Find the height of the building correct to the nearest metre.
Answer: Let AB be the height of the building, \( h \) metres. Let the man initially be at point D, where the angle of elevation to the top of the building (A) is 30° (\( \angle ADB = 30° \)). He then walks 60 m towards the building to point C, where the angle of elevation changes to 60° (\( \angle ACB = 60° \)). So, CD = 60 m. Let BC be \( x \) metres.
In the right-angled triangle \( \triangle ACB \):
\( \tan 60° = \frac{AB}{BC} \)
\( \sqrt{3} = \frac{h}{x} \)
\( x = \frac{h}{\sqrt{3}} \) (Equation 1)
In the right-angled triangle \( \triangle ADB \):
\( \tan 30° = \frac{AB}{BD} \)
\( \frac{1}{\sqrt{3}} = \frac{h}{BC + CD} \)
\( \frac{1}{\sqrt{3}} = \frac{h}{x + 60} \)
Substitute \( x \) from Equation 1 into this equation:
\( \frac{1}{\sqrt{3}} = \frac{h}{\frac{h}{\sqrt{3}} + 60} \)
Cross-multiply:
\( \frac{h}{\sqrt{3}} + 60 = h\sqrt{3} \)
\( 60 = h\sqrt{3} - \frac{h}{\sqrt{3}} \)
\( 60 = h (\sqrt{3} - \frac{1}{\sqrt{3}}) \)
\( 60 = h (\frac{3-1}{\sqrt{3}}) \)
\( 60 = h (\frac{2}{\sqrt{3}}) \)
\( h = \frac{60\sqrt{3}}{2} \)
\( h = 30\sqrt{3} \)
Using \( \sqrt{3} \approx 1.732 \):
\( h = 30 \times 1.732 \)
\( h = 51.96 \) m
Rounding to the nearest metre, the height of the building is 52 m. Moving closer to an object increases its angle of elevation.
In simple words: A person sees the top of a building at a 30-degree angle. After walking 60 metres closer, the angle becomes 60 degrees. By using these angles, we can figure out that the building is about 52 metres tall.

🎯 Exam Tip: Always make sure your diagram accurately represents the scenario, especially the order of points when moving towards or away from an object. The larger angle of elevation should always correspond to the closer observation point.

Question 14. As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships on the same side of the lighthouse in horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest metre.
Answer: Let AB be the lighthouse, with height 80 m. Let D and C be the positions of the two ships on the same side of the lighthouse's base (B). The ship D is further away, and C is closer. The angle of depression from the top of the lighthouse (A) to ship D is 30°, so \( \angle ADB = 30° \) (alternate angle). The angle of depression from A to ship C is 40°, so \( \angle ACB = 40° \) (alternate angle). We need to find the distance between the ships, which is CD = BD - BC.
In the right-angled triangle \( \triangle ACB \):
\( \tan 40° = \frac{AB}{BC} \)
\( BC = \frac{80}{\tan 40°} \)
\( BC = \frac{80}{0.8391} \) (Using \( \tan 40° \approx 0.8391 \))
\( BC = 95.339 \) m
In the right-angled triangle \( \triangle ADB \):
\( \tan 30° = \frac{AB}{BD} \)
\( BD = \frac{80}{\tan 30°} \)
\( BD = 80\sqrt{3} \)
Using \( \sqrt{3} \approx 1.732 \):
\( BD = 80 \times 1.732 = 138.56 \) m
The distance between the two ships is CD = BD - BC.
\( CD = 138.56 - 95.339 = 43.221 \) m
Rounding to the nearest metre, the distance between the ships is 43 m. A larger angle of depression means the object is closer.
In simple words: From the top of an 80-metre lighthouse, two ships are seen on the same side. The further ship is seen at a 30-degree downward angle, and the closer one at a 40-degree downward angle. By calculating their distances from the lighthouse, we find the ships are about 43 metres apart.

🎯 Exam Tip: When two objects are on the same side, the distance between them is the absolute difference of their individual distances from the base. Always draw the diagram carefully, placing the larger angle of depression for the closer object.

Question 15. In the figure given, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find: (i) the horizontal distance between AB and CD. (ii) the height of the lamp post.
Answer: Let AB be the building, with height 60 m. Let CD be the lamp post, with height \( h \) metres. Let the horizontal distance between the building and the lamp post be BD = \( x \) metres.
Draw a horizontal line CE from C (top of the lamp post) to meet the building at E. This creates a rectangle CEDB. So, CE = BD = \( x \) and EB = CD = \( h \). The height AE = AB - EB = \( (60 - h) \) metres.
The angle of depression from the top of the building (A) to the bottom of the lamp post (D) is 60°. So, in the right-angled triangle \( \triangle ABD \):
\( \tan 60° = \frac{AB}{BD} \)
\( \sqrt{3} = \frac{60}{x} \)
\( x = \frac{60}{\sqrt{3}} \)
\( x = \frac{60\sqrt{3}}{3} = 20\sqrt{3} \) metres
Using \( \sqrt{3} \approx 1.732 \):
\( x = 20 \times 1.732 = 34.64 \) m
(i) Therefore, the horizontal distance between AB and CD is 34.64 m.
The angle of depression from the top of the building (A) to the top of the lamp post (C) is 30°. So, in the right-angled triangle \( \triangle AEC \):
\( \tan 30° = \frac{AE}{CE} \)
\( \frac{1}{\sqrt{3}} = \frac{60-h}{x} \)
Substitute \( x = 20\sqrt{3} \) metres:
\( \frac{1}{\sqrt{3}} = \frac{60-h}{20\sqrt{3}} \)
Multiply both sides by \( 20\sqrt{3} \):
\( 20\sqrt{3} \times \frac{1}{\sqrt{3}} = 60-h \)
\( 20 = 60-h \)
\( h = 60-20 \)
\( h = 40 \) m
(ii) Therefore, the height of the lamp post is 40 m. Setting up a rectangle with a horizontal line simplifies calculations involving two vertical objects.
In simple words: From the top of a 60-metre building, we look at a lamp post. The angle looking down to the lamp post's top is 30 degrees, and to its bottom is 60 degrees. First, we find the ground distance between the building and the lamp post, which is 34.64 metres. Then, we find that the lamp post is 40 metres tall.

🎯 Exam Tip: For problems involving two vertical objects where observations are made from one to the other, always draw a horizontal line from the observation point to the other object. This creates a right triangle and a rectangle, helping to relate the heights and distances.

Question 16. An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number.
Answer: Let O be the position of the aeroplane, and M be the point on the river directly below the aeroplane. So, the altitude OM = 250 m. Let A and B be the positions of the two boats on opposite banks of the river. The angles of depression from O to A and B are 45° and 60° respectively. This means the angles of elevation from the boats to the aeroplane are \( \angle OAM = 45° \) and \( \angle OBM = 60° \) (alternate interior angles). The width of the river is AB = AM + MB.
In the right-angled triangle \( \triangle OMA \):
\( \tan 45° = \frac{OM}{AM} \)
\( 1 = \frac{250}{AM} \)
\( AM = 250 \) m
In the right-angled triangle \( \triangle OMB \):
\( \tan 60° = \frac{OM}{BM} \)
\( \sqrt{3} = \frac{250}{BM} \)
\( BM = \frac{250}{\sqrt{3}} \)
Using \( \sqrt{3} \approx 1.732 \):
\( BM = \frac{250}{1.732} = 144.3418... \) m
The total width of the river is AM + BM.
\( AB = 250 + 144.3418 = 394.3418 \) m
Rounding to the nearest whole number, the width of the river is 394 m. The tangent of 45 degrees being 1 simplifies one part of the calculation.
In simple words: An aeroplane is 250 metres high above a river. It sees two boats on opposite sides. One boat is seen at a 45-degree downward angle, and the other at a 60-degree downward angle. We calculate the distance of each boat from the point directly under the plane and add them up. The river is about 394 metres wide.

🎯 Exam Tip: When objects are on opposite sides, the total distance is the sum of their individual distances from the point directly below the observation point. Remember to round your final answer as specified (e.g., to the nearest whole number).

Question 17. The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is 30° and 24° respectively. Find the height of the two towers. Give your answer correct to 3 significant figures.
Answer: Let the first tower be AB with height \( h_1 \), and the second tower be CD with height \( h_2 \). The horizontal distance between them, BD, is 120 m. Observations are made from the top of the second tower (C).
Draw a horizontal line CE from C to meet the first tower AB at E. This creates a rectangle CEDB. So, CE = BD = 120 m and EB = CD = \( h_2 \). The height AE = AB - EB = \( (h_1 - h_2) \) m.
1. The angle of elevation of the top of the first tower (A) from C is 30°. So, in the right-angled triangle \( \triangle AEC \):
\( \tan 30° = \frac{AE}{CE} \)
\( \frac{1}{\sqrt{3}} = \frac{AE}{120} \)
\( AE = \frac{120}{\sqrt{3}} \)
\( AE = \frac{120\sqrt{3}}{3} = 40\sqrt{3} \) m
Using \( \sqrt{3} \approx 1.732 \):
\( AE = 40 \times 1.732 = 69.28 \) m
2. The angle of depression of the bottom of the first tower (B) from C is 24°. So, in the right-angled triangle \( \triangle CEB \):
\( \tan 24° = \frac{EB}{CE} \)
Using \( \tan 24° \approx 0.4452 \):
\( 0.4452 = \frac{EB}{120} \)
\( EB = 0.4452 \times 120 = 53.424 \) m
Now we find the heights of the towers:
The height of the second tower CD = EB = 53.424 m.
The height of the first tower AB = AE + EB = \( 69.28 + 53.424 = 122.704 \) m.
Rounding to 3 significant figures:
Height of the first tower \( h_1 \approx 123 \) m.
Height of the second tower \( h_2 \approx 53.4 \) m. Drawing a horizontal line from the observation point helps set up the trigonometric equations.
In simple words: Two towers are 120 metres apart. From the top of the second tower, looking up to the first tower's top is a 30-degree angle, and looking down to its bottom is a 24-degree angle. The first tower is about 123 metres tall, and the second tower is about 53.4 metres tall.

🎯 Exam Tip: When dealing with two towers of different heights, always draw a horizontal line from the observation point (usually the top of one tower) to create two right-angled triangles. This helps solve for the unknown parts efficiently.

Question 18. An aeroplane at an altitude of 1500 metres finds that two ships are sailing towards it in the same direction. The angles of depression as observed from the aeroplane are 45° and 30° respectively. Find the distance between the two ships.
Answer: Let A be the position of the aeroplane, and B be the point on the ground directly below it. So, the altitude AB = 1500 m. Let C and D be the positions of the two ships sailing in the same direction towards the aeroplane. Ship C is closer, and ship D is further away. The angles of depression from the aeroplane (A) to ship C and ship D are 45° and 30° respectively. This means the angles of elevation from the ships to the aeroplane are \( \angle ACB = 45° \) and \( \angle ADB = 30° \) (alternate interior angles). We need to find the distance between the two ships, which is CD = BD - BC.
In the right-angled triangle \( \triangle ABC \):
\( \tan 45° = \frac{AB}{BC} \)
\( 1 = \frac{1500}{BC} \)
\( BC = 1500 \) m
In the right-angled triangle \( \triangle ABD \):
\( \tan 30° = \frac{AB}{BD} \)
\( \frac{1}{\sqrt{3}} = \frac{1500}{BD} \)
\( BD = 1500\sqrt{3} \) m
Using \( \sqrt{3} \approx 1.732 \):
\( BD = 1500 \times 1.732 = 2598 \) m
The distance between the two ships is CD = BD - BC.
\( CD = 2598 - 1500 = 1098 \) m
The ship further away will always have a smaller angle of depression.
In simple words: An aeroplane is 1500 metres high. It sees two ships below, sailing the same way. The closer ship is seen at a 45-degree downward angle, and the further ship at a 30-degree downward angle. We calculate that the two ships are 1098 metres apart.

🎯 Exam Tip: When objects are on the same side and you're finding the distance between them, calculate each object's distance from the base of the observation point and then find the difference. Always use the correct angle for each calculation.