CBSE Class 10 Mathematics Applications of Trigonometry Assignment Set 03

KEY POINTS

Line of sight: Line segment joining the object to the eye of the observer is called the line of sight.

Angle of elevation: When an observer sees an object situated in upward direction, the angle formed by line of sight with horizontal line is called angle of elevation.

Angle of depression: When an observer sees an object situated in downward direction the angle formed by line of sight with horizontal line is called angle of depression.

LEVEL - 1

Question 1. A 6cm high casts a shadow \(2\sqrt{3}\text{m}\) long on the ground, then find the sun's elevation?
Answer: Let height of the pole be \(h = 6\text{m}\) (assuming height is in meters as per standard shadow problems, or directly use \(6\text{cm}\) and shadow \(2\sqrt{3}\text{m}\)). Let the sun's elevation be \(\theta\).
Using trigonometry in right-angled triangle:
\(\tan\theta = \frac{\text{Height}}{\text{Shadow}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}\)
\(\tan\theta = \sqrt{3} \Rightarrow \theta = 60^{\circ}\)
In simple words: The ratio of height to shadow gives the tangent of the elevation angle, which is \(\sqrt{3}\), corresponding to \(60^{\circ}\).

Exam Tip: Remember standard trigonometric values: \(\tan 60^{\circ} = \sqrt{3}\).

Question 2. If \(\sqrt{3}\tan\theta = 1\), then find the value of \(\sin^2\theta - \cos^2\theta\)
Answer: Given \(\sqrt{3}\tan\theta = 1 \Rightarrow \tan\theta = \frac{1}{\sqrt{3}}\)
Since \(\tan 30^{\circ} = \frac{1}{\sqrt{3}}\), we get \(\theta = 30^{\circ}\).
Now, find the value of \(\sin^2\theta - \cos^2\theta\):
\(\sin 30^{\circ} = \frac{1}{2}\) and \(\cos 30^{\circ} = \frac{\sqrt{3}}{2}\)
\(\sin^2 30^{\circ} - \cos^2 30^{\circ} = \left(\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} - \frac{3}{4} = \frac{-2}{4} = -\frac{1}{2}\)
In simple words: First find angle \(\theta = 30^{\circ}\) from the given equation, then substitute the values of sine and cosine.

Exam Tip: Be careful with signs when subtracting squared trigonometric ratios.

Question 3. An observer 1.5m tall is 20.5 metres away from a tower 22m high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Answer: Height of the tower = \(22\text{m}\)
Height of the observer = \(1.5\text{m}\)
Height of the tower above the observer's eye level = \(22 - 1.5 = 20.5\text{m}\)
Distance of the observer from the tower = \(20.5\text{m}\)
Let \(\theta\) be the angle of elevation. Then:
\(\tan\theta = \frac{20.5}{20.5} = 1\)
\(\theta = 45^{\circ}\)
In simple words: Subtract the observer's height from the tower's total height to get the vertical height from the eye level, which equals the horizontal distance.

Exam Tip: Always subtract observer's eye height from total height when given.

Question 4. A ladder 15m long just reaches the top of vertical wall. If the ladder makes an angle \(60^{\circ}\) with the wall, find the height of the wall
Answer: Length of the ladder (hypotenuse) \(l = 15\text{m}\)
Angle made by the ladder with the wall = \(60^{\circ}\). Therefore, angle made with the ground is \(90^{\circ} - 60^{\circ} = 30^{\circ}\), or using triangle rules directly with the wall angle: let height of wall be \(h\).
\(\cos(60^{\circ}) = \frac{\text{Height of the wall}}{\text{Length of the ladder}} = \frac{h}{15}\)
Since \(\cos(60^{\circ}) = \frac{1}{2}\):
\(\frac{h}{15} = \frac{1}{2} \Rightarrow h = \frac{15}{2}\text{m}\)
In simple words: Since the angle is given with the wall, use cosine to find the adjacent side which is the wall's height.

Exam Tip: Note whether the angle is given with the vertical wall or horizontal ground.

Question 5. In a rectangle ABCD, AB = 20cm \(\angle BAC = 60^{\circ}\) then find the length of the side AD.
Answer: In rectangle ABCD, \(\angle B = 90^{\circ}\), so \(\triangle ABC\) is a right-angled triangle right-angled at B.
Given \(AB = 20\text{cm}\) and \(\angle BAC = 60^{\circ}\).
Side AD is equal to side BC in a rectangle.
Using \(\tan(\angle BAC) = \frac{\text{BC}}{\text{AB}}\):
\(\tan(60^{\circ}) = \frac{\text{BC}}{20}\)
\(\sqrt{3} = \frac{\text{BC}}{20} \Rightarrow \text{BC} = 20\sqrt{3}\text{cm}\)
Since \(AD = BC\), length of side \(AD = 20\sqrt{3}\text{cm}\).
In simple words: Use the tangent ratio in right-angled triangle ABC to find BC, which is equal to AD.

Exam Tip: Opposite sides of a rectangle are equal in length.

Question 6. Find the angle of elevation of the sun's altitude when the height of the shadow of a vertical pole is equal to its height:
Answer: Let the height of the pole be \(h\) and the length of its shadow be \(s\).
Given that height equals shadow: \(h = s\).
Let \(\theta\) be the angle of elevation. Then:
\(\tan\theta = \frac{h}{s} = \frac{h}{h} = 1\)
\(\theta = 45^{\circ}\)
In simple words: When height and shadow are equal, the angle is always \(45^{\circ}\) because \(\tan 45^{\circ} = 1\).

Exam Tip: Equal height and shadow always imply an angle of elevation of \(45^{\circ}\).

Question 7. From a point 20m away from the foot of a tower, the angle of elevation of top of the tower is \(30^{\circ}\), find the height of the tower.
Answer: Let the height of the tower be \(h\).
Distance from the foot = \(20\text{m}\), angle of elevation \(\theta = 30^{\circ}\).
\(\tan(30^{\circ}) = \frac{h}{20}\)
\(\frac{1}{\sqrt{3}} = \frac{h}{20} \Rightarrow h = \frac{20}{\sqrt{3}}\text{m}\)
In simple words: Use the tangent ratio with distance and height to solve for \(h\).

Question 8. In the adjacent figure, what are the angles of depression of the top and bottom of a pole from the top of a tower \(h\) m high:
Answer: Based on the geometry of the given figure (with angles \(45^{\circ}\) and \(30^{\circ}\) marked from the top Q of the tower to the top A and bottom B of the pole):
Angle of depression of the top of the pole = \(45^{\circ}\)
Angle of depression of the bottom of the pole = \(30^{\circ}\)
In simple words: The alternate interior angles show that the angles of depression from the top of the tower correspond to \(45^{\circ}\) and \(30^{\circ}\).

Exam Tip: Angle of elevation equals angle of depression due to alternate interior angles.

LEVEL - 2

Question 1. In \(\triangle ABC\), \(\angle B = 45^{\circ}\), \(\angle C = 45^{\circ}\), \(AB = 5\text{cm}\) then find the length of the other two sides.
Answer: Since \(\angle B = \angle C = 45^{\circ}\), the triangle is an isosceles right-angled triangle where \(AB = AC = 5\text{cm}\).
The third side (hypotenuse) \(BC\) can be found using Pythagoras theorem:
\(BC^2 = AB^2 + AC^2 = 5^2 + 5^2 = 25 + 25 = 50\)
\(BC = \sqrt{50} = 5\sqrt{2}\text{cm}\)
Thus, the other two sides are \(5\text{cm}\) and \(5\sqrt{2}\text{cm}\).
In simple words: Equal angles mean equal opposite sides, and the hypotenuse is found using Pythagoras theorem.

Exam Tip: In a \(45^{\circ}-45^{\circ}-90^{\circ}\) triangle, hypotenuse = side \(\times \sqrt{2}\).

Question 2. From a point 20m away from the foot of the tower, the angle of elevation of the top of the tower is \(30^{\circ}\), find the height of the tower.
Answer: Let the height of the tower be \(h\).
\(\tan(30^{\circ}) = \frac{h}{20}\)
\(\frac{1}{\sqrt{3}} = \frac{h}{20} \Rightarrow h = \frac{20}{\sqrt{3}}\text{m} = \frac{20\sqrt{3}}{3}\text{m}\)
In simple words: Multiply both sides by 20 and substitute \(\tan 30^{\circ}\).

Exam Tip: Rationalized form \(\frac{20\sqrt{3}}{3}\text{m}\) is often preferred in final answers.

Question 3. A ladder 50m long just reaches the top of a vertical wall. If the ladder makes an angle of \(60^{\circ}\) with the wall, find the height of the wall.
Answer: Let height of wall be \(h\), ladder length \(l = 50\text{m}\).
Angle with the wall = \(60^{\circ}\).
\(\cos(60^{\circ}) = \frac{h}{50}\)
\(\frac{1}{2} = \frac{h}{50} \Rightarrow h = 25\text{m}\)
In simple words: Use cosine of the angle with the vertical wall.

Exam Tip: Check carefully whether the angle is given with the ground or the wall.

Question 4. A circus artist is climbing a 20m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is \(30^{\circ}\).
Answer: Let height of pole be \(h\). Length of rope (hypotenuse) = \(20\text{m}\), angle \(\theta = 30^{\circ}\).
\(\sin(30^{\circ}) = \frac{\text{Height}}{\text{Length of rope}} = \frac{h}{20}\)
\(\frac{1}{2} = \frac{h}{20} \Rightarrow h = 10\text{m}\)
In simple words: Use the sine ratio since we know the hypotenuse and need the opposite side.

Exam Tip: Rope length acts as the hypotenuse in right triangle problems.

Question 5. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle \(30^{\circ}\) with it. The distance between the foot of the tree to the point where the top touches the ground is 8m. Find the height of the tree.
Answer: Let the remaining height of the tree be \(h\) and the broken part be \(x\) (which acts as the hypotenuse).
Distance from foot to top on ground = \(8\text{m}\).
Using \(\tan(30^{\circ})\) for the vertical height \(h\):
\(\tan(30^{\circ}) = \frac{h}{8} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{8} \Rightarrow h = \frac{8}{\sqrt{3}}\text{m}\)
Using \(\cos(30^{\circ})\) for the broken part \(x\):
\(\cos(30^{\circ}) = \frac{8}{x} \Rightarrow \frac{\sqrt{3}}{2} = \frac{8}{x} \Rightarrow x = \frac{16}{\sqrt{3}}\text{m}\)
Total height of the tree = \(h + x = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3}\text{m}\).
In simple words: Total height is the sum of the standing part and the broken slanted part.

Exam Tip: Do not forget to add both parts (standing height + broken hypotenuse) for the total height of the tree.

LEVEL - 3

Question 1. The shadow of a tower standing on a level plane is found to be 50m longer when sun's elevation is \(30^{\circ}\) then when it is \(60^{\circ}\). Find the height of the tower.
Answer: Let height of tower be \(h\) and shadow length for \(60^{\circ}\) be \(x\).
When elevation is \(60^{\circ}\): \(\tan(60^{\circ}) = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}\).
When elevation is \(30^{\circ}\), shadow is \(x + 50\):
\(\tan(30^{\circ}) = \frac{h}{x + 50} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{\frac{h}{\sqrt{3}} + 50}\)
\(\frac{h}{\sqrt{3}} + 50 = h\sqrt{3}\)
\(50 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h\left(\frac{3 - 1}{\sqrt{3}}\right) = \frac{2h}{\sqrt{3}}\)
\(2h = 50\sqrt{3} \Rightarrow h = 25\sqrt{3}\text{m}\)
In simple words: Set up two equations using tangent for both angles and solve simultaneously for height \(h\).

Exam Tip: Standard two-position shadow problem: \(h = d \left(\frac{\tan\theta_1 \tan\theta_2}{\tan\theta_2 - \tan\theta_1}\right)\).

Question 2. The angle of depression of the top and bottom of a tower as seen from the top of a 100m high cliff are \(30^{\circ}\) and \(60^{\circ}\) respectively. Find the height of the tower.
Answer: Let height of the cliff be \(CD = 100\text{m}\) and height of the tower be \(AB = h\).
Let distance between cliff and tower be \(x\).
From top of cliff C, angle of depression of bottom of tower B is \(60^{\circ}\), so angle of elevation from B to C is \(60^{\circ}\):
\(\tan(60^{\circ}) = \frac{100}{x} \Rightarrow \sqrt{3} = \frac{100}{x} \Rightarrow x = \frac{100}{\sqrt{3}}\text{m}\)
Angle of depression of top of tower A is \(30^{\circ}\), so height of cliff above tower top is \(100 - h\):
\(\tan(30^{\circ}) = \frac{100 - h}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100 - h}{\frac{100}{\sqrt{3}}}\)
\(100 - h = \frac{100}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = \frac{100}{3}\)
\(h = 100 - \frac{100}{3} = \frac{200}{3} = 66.67\text{m}\)
In simple words: Use the distance found from the bottom angle to calculate the height difference for the top angle.

Question 3. From a window (9m above ground) of a house in a street, the angles of elevation and depression of the top and foot of another house on the opposite side of the street are \(30^{\circ}\) and \(60^{\circ}\) respectively. Find the height of the opposite house and width of the street.
Answer: Let window be at height \(h_1 = 9\text{m}\).
Width of the street = \(x\).
Angle of depression of foot of opposite house is \(60^{\circ}\):
\(\tan(60^{\circ}) = \frac{9}{x} \Rightarrow \sqrt{3} = \frac{9}{x} \Rightarrow x = \frac{9}{\sqrt{3}} = 3\sqrt{3}\text{m} \approx 5.2\text{m}\) (Width of the street)
Angle of elevation of top of opposite house is \(30^{\circ}\). Let height above window be \(h_2\):
\(\tan(30^{\circ}) = \frac{h_2}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h_2}{3\sqrt{3}} \Rightarrow h_2 = 3\text{m}\)
Total height of opposite house = \(9 + h_2 = 9 + 3 = 12\text{m}\).
In simple words: The lower triangle gives the street width, and the upper triangle gives the height of the opposite house above the window level.

Exam Tip: Add the window height to the upper calculated height to get the total height.

Question 4. From the top of a hill, the angle of depression of two consecutive kilometer stones due east are found to be \(30^{\circ}\) and \(45^{\circ}\). Find the height of the hill.
Answer: Let height of hill be \(h\) and distance of first milestone be \(x\). Distance between consecutive milestone stones is \(1\text{km}\).
\(\tan(45^{\circ}) = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \Rightarrow x = h\)
\(\tan(30^{\circ}) = \frac{h}{x + 1} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{h + 1}\)
\(h + 1 = h\sqrt{3} \Rightarrow h(\sqrt{3} - 1) = 1\)
\(h = \frac{1}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{2} \approx \frac{1.732 + 1}{2} = 1.366\text{km} \approx 1.37\text{km}\)
In simple words: Consecutive stones are 1km apart. Use tangent equations for both angles to solve for height.

Exam Tip: Rationalize the denominator using \((\sqrt{3} + 1)\) to simplify numerical evaluation.

Question 5. Two poles of equal heights are standing opposite each other on either side of the road, which is 80m wide. From a point between them on the road the angles of elevation of the top of the poles are \(60^{\circ}\) and \(30^{\circ}\). Find the heights of pole and the distance of the point from the poles.
Answer: Let height of both poles be \(h\). Total width = \(80\text{m}\).
Let distance of point from first pole be \(x\), so distance from second pole is \(80 - x\).
\(\tan(60^{\circ}) = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}\)
\(\tan(30^{\circ}) = \frac{h}{80 - x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{80 - x} \Rightarrow h = \frac{80 - x}{\sqrt{3}}\)
Equating both expressions for \(h\):
\(x\sqrt{3} = \frac{80 - x}{\sqrt{3}}\)
\(3x = 80 - x \Rightarrow 4x = 80 \Rightarrow x = 20\text{m}\)
Distance from second pole = \(80 - 20 = 60\text{m}\).
Height \(h = 20\sqrt{3} = 20 \times 1.732 = 34.64\text{m}\).
In simple words: Split the 80m road into \(x\) and \(80 - x\), equate the height expressions, and solve for \(x\).

Exam Tip: The point is closer to the pole with the larger angle of elevation (\(60^{\circ}\)).

Question 6. The angle of elevation of a jet fighter from a point A on the ground is \(60^{\circ}\). After a flight of 15 seconds, the angle of elevation changes to \(30^{\circ}\). If the jet is flying at a speed of \(720\text{km/hr}\), find the constant height at which the jet is flying.
Answer: Speed of jet = \(720\text{km/hr} = 720 \times \frac{5}{18} = 200\text{m/s}\).
Distance traveled in 15 seconds = \(200 \times 15 = 3000\text{m}\).
Let constant height be \(h\).
Initially, distance from observation point A = \(\frac{h}{\tan(60^{\circ})} = \frac{h}{\sqrt{3}}\).
After 15 seconds, total horizontal distance = \(\frac{h}{\tan(30^{\circ})} = h\sqrt{3}\).
Distance traveled = \(h\sqrt{3} - \frac{h}{\sqrt{3}} = 3000\text{m}\)
\(h\left(\frac{3 - 1}{\sqrt{3}}\right) = 3000 \Rightarrow \frac{2h}{\sqrt{3}} = 3000 \Rightarrow 2h = 3000\sqrt{3} \Rightarrow h = 1500\sqrt{3}\text{m}\) (or \(1500\text{m}\) if distance is interpreted appropriately based on standard textbook values where speed/distance yields \(1500\text{m}\)).
In simple words: Calculate distance covered by speed \(\times\) time, then use horizontal ground distances corresponding to both elevation angles.

Exam Tip: Convert speed from km/hr to m/s by multiplying by \(\frac{5}{18}\).

Question 7. A window in a building is at a height of 10m above the ground. The angle of depression of a point P on the ground from the window is \(30^{\circ}\). The angle of elevation of the top of the building from the point P is \(60^{\circ}\). Find the height of the building.
Answer: Height of window = \(10\text{m}\).
Distance of point P from the building base, let it be \(x\):
From the window triangle: \(\tan(30^{\circ}) = \frac{10}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{10}{x} \Rightarrow x = 10\sqrt{3}\text{m}\).
Let total height of the building be \(H\).
From point P to the top of the building with angle \(60^{\circ}\):
\(\tan(60^{\circ}) = \frac{H}{x} \Rightarrow \sqrt{3} = \frac{H}{10\sqrt{3}} \Rightarrow H = 10\sqrt{3} \times \sqrt{3} = 30\text{m}\).
In simple words: First find the distance to point P using the window height, then use that distance to find the total height of the building.

Exam Tip: Two right triangles sharing the same horizontal base make these problems easy to solve sequentially.

Question 8. A boy, whose eye level is 1.3m from the ground, spots a balloon moving with the wind in a horizontal line at same height from the ground. The angle of elevation of the balloon from the eyes of the boy at any instant is \(60^{\circ}\). After 2 seconds, the angle of elevation reduces to \(30^{\circ}\). If the speed of the wind at that moment is \(29\sqrt{3}\text{m/s}\), then find the height of the balloon from the ground.
Answer: Speed of wind = \(29\sqrt{3}\text{m/s}\), time = \(2\text{seconds}\).
Distance traveled by balloon = \(29\sqrt{3} \times 2 = 58\sqrt{3}\text{m}\).
Let height of balloon above eye level be \(h\).
Initial horizontal distance = \(\frac{h}{\tan(60^{\circ})} = \frac{h}{\sqrt{3}}\).
Final horizontal distance = \(\frac{h}{\tan(30^{\circ})} = h\sqrt{3}\).
Distance traveled = \(h\sqrt{3} - \frac{h}{\sqrt{3}} = 58\sqrt{3}\)
\(\frac{2h}{\sqrt{3}} = 58\sqrt{3} \Rightarrow 2h = 58 \times 3 = 174 \Rightarrow h = 87\text{m}\).
Total height of balloon from ground = height above eye level + eye level = \(87 + 1.3 = 88.3\text{m}\).
In simple words: Subtract eye level from the final height calculation and find the height above the eyes first.

Exam Tip: Always remember to add the observer's eye height back to get the final height from the ground.

Question 9. A man on the deck on a ship 14m above water level, observes that the angle of elevation of the top of a cliff is \(60^{\circ}\) and the angle of depression of the base of the cliff is \(30^{\circ}\). Calculate the distance of the cliff from the ship and the height of the cliff.
Answer: Height of deck = \(14\text{m}\).
Angle of depression of base of cliff = \(30^{\circ}\), so distance \(x\) from ship to cliff is given by:
\(\tan(30^{\circ}) = \frac{14}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{14}{x} \Rightarrow x = 14\sqrt{3} = 14 \times 1.732 = 24.25\text{m}\) (Distance).
Angle of elevation of top of cliff = \(60^{\circ}\). Let height of cliff above deck be \(h_2\):
\(\tan(60^{\circ}) = \frac{h_2}{14\sqrt{3}} \Rightarrow \sqrt{3} = \frac{h_2}{14\sqrt{3}} \Rightarrow h_2 = 14\sqrt{3} \times \sqrt{3} = 42\text{m}\).
Total height of cliff = \(14 + 42 = 56\text{m}\).
In simple words: Use the deck height and depression angle to find the distance, then use that distance and elevation angle to find the upper height of the cliff.

Exam Tip: Total height of the cliff is the sum of the deck height and the height above the deck.

Question 10. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of \(30^{\circ}\), which is approaching the foot of tower with a uniform speed. Six minutes later, the angle of depression of the car is found to be \(60^{\circ}\). Find the time taken by the car to reach the foot of the tower.
Answer: Let height of tower be \(h\).
Let speed of car be \(v\). Distance covered in 6 minutes = \(6v\).
When angle of depression is \(60^{\circ}\), distance from tower \(x = \frac{h}{\tan(60^{\circ})} = \frac{h}{\sqrt{3}}\).
When angle of depression is \(30^{\circ}\), total distance = \(\frac{h}{\tan(30^{\circ})} = h\sqrt{3}\).
Distance traveled in 6 minutes = \(h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}\).
Remaining distance to reach foot = \(\frac{h}{\sqrt{3}}\), which is exactly half of the distance covered in 6 minutes.
Therefore, time taken = \(\frac{6\text{ minutes}}{2} = 3\text{minutes}\).
In simple words: The distance for the second half is half the distance of the first part, so it takes half the time (3 minutes).

Exam Tip: For uniform speed, time taken is directly proportional to distance.

SELF EVALUATION / HOTS

Question 1. An aeroplane when flying at a height of 3125m from the ground passes vertically below another plane at an instant when the angle of elevation of the two planes from the same point on the ground are \(30^{\circ}\) and \(60^{\circ}\) respectively. Find the distance between the two planes at that instant.
Answer: Height of first plane = \(3125\text{m}\). Let height of second plane be \(H\).
Distance from point on ground = \(x\).
For first plane (\(30^{\circ}\)): \(\tan(30^{\circ}) = \frac{3125}{x} \Rightarrow x = 3125\sqrt{3}\text{m}\).
For second plane (\(60^{\circ}\)): \(\tan(60^{\circ}) = \frac{H}{x} \Rightarrow \sqrt{3} = \frac{H}{3125\sqrt{3}} \Rightarrow H = 3125\sqrt{3} \times \sqrt{3} = 3125 \times 3 = 9375\text{m}\).
Distance between the two planes = \(H - 3125 = 9375 - 3125 = 6250\text{m}\).
In simple words: Find the horizontal base distance using the lower plane, then calculate the total height of the upper plane using the higher angle of elevation.

Exam Tip: The horizontal base distance remains the same for both planes from the same observation point.

Question 2. From the top of a building 60m high, the angles of depression of the top and bottom of a vertical lamp post are observed to be \(30^{\circ}\) and \(60^{\circ}\) respectively. Find [i] horizontal distance between the building and the lamp post [ii] height of the lamp post.
Answer: Height of building = \(60\text{m}\).
[ii] Height of the lamp post let be \(h\), and [i] horizontal distance be \(x\).
Angle of depression of bottom of lamp post is \(60^{\circ}\):
\(\tan(60^{\circ}) = \frac{60}{x} \Rightarrow \sqrt{3} = \frac{60}{x} \Rightarrow x = \frac{60}{\sqrt{3}} = 20\sqrt{3} = 20 \times 1.732 = 34.64\text{m}\) (Horizontal distance).
Angle of depression of top of lamp post is \(30^{\circ}\):
Height difference = \(60 - h\).
\(\tan(30^{\circ}) = \frac{60 - h}{20\sqrt{3}} \Rightarrow \frac{1}{\sqrt{3}} = \frac{60 - h}{20\sqrt{3}} \Rightarrow 60 - h = 20 \Rightarrow h = 40\text{m}\) (Height of lamp post).
In simple words: Use bottom depression to find distance, then use top depression to find the remaining height difference.

Exam Tip: Always calculate the horizontal distance first as it connects both triangles.

Question 3. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height \(h\) m. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are \(\alpha\) and \(\beta\), respectively. Prove that the height of the tower is \(\frac{h\tan\alpha}{\tan\beta - \tan\alpha}\).
Answer: Let the height of the tower be \(H\) and the distance of the point from the base of the tower be \(x\).
Angle of elevation of the bottom of the flag staff (top of tower) = \(\alpha\):
\(\tan\alpha = \frac{H}{x} \Rightarrow x = \frac{H}{\tan\alpha}\)
Angle of elevation of the top of the flag staff = \(\beta\):
\(\tan\beta = \frac{H + h}{x} \Rightarrow x = \frac{H + h}{\tan\beta}\)
Equating both expressions for \(x\):
\(\frac{H}{\tan\alpha} = \frac{H + h}{\tan\beta}\)
\(H\tan\beta = (H + h)\tan\alpha\)
\(H\tan\beta = H\tan\alpha + h\tan\alpha\)
\(H(\tan\beta - \tan\alpha) = h\tan\alpha\)
\(H = \frac{h\tan\alpha}{\tan\beta - \tan\alpha} \quad \text{[Hence Proved]}\)
In simple words: Equate the horizontal distance \(x\) obtained from both triangles and rearrange terms to isolate \(H\).

Exam Tip: This is a standard algebraic proof in heights and distances; memorize the substitution steps.

Question 4. The angle of elevation of a cloud from a point 60m above a lake is \(30^{\circ}\) and the angle of depression of the reflection of the cloud in the lake is \(60^{\circ}\). Find the height of the cloud from the surface of the lake.
Answer: Let the height of the cloud above the lake be \(H\).
Point of observation is \(60\text{m}\) above the lake.
Height of cloud above the observation point = \(H - 60\).
Depth of the reflection of the cloud below the lake surface = \(H + 60\).
Let horizontal distance from observation point to cloud be \(x\).
From observation point to cloud (\(30^{\circ}\)):
\(\tan(30^{\circ}) = \frac{H - 60}{x} \Rightarrow x = \frac{H - 60}{\tan(30^{\circ})} = (H - 60)\sqrt{3}\)
From observation point to the reflection of cloud (\(60^{\circ}\)):
\(\tan(60^{\circ}) = \frac{H + 60}{x} \Rightarrow x = \frac{H + 60}{\tan(60^{\circ})} = \frac{H + 60}{\sqrt{3}}\)
Equating both expressions for \(x\):
\((H - 60)\sqrt{3} = \frac{H + 60}{\sqrt{3}}\)
\(3(H - 60) = H + 60\)
\(3H - 180 = H + 60\)
\(2H = 240 \Rightarrow H = 120\text{m}\)
In simple words: The height of the cloud above water is \(H\), its reflection is at \(-H\), and equating the horizontal distances yields the height.

Exam Tip: The reflection of an object in water is at the same distance below the water surface as the object is above it.