Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 22 Heights And Distances

Exercise 22.1

Question 2. Find the height of a tower if the angle of elevation from a point 30 m away from its base is 30°.
Answer: Let AB be the cliff and angle of elevation from point C (on ground) is 30°.

In \( \triangle ABC \)

\( \frac{AB}{BC} = \tan 60° \)
\( \frac{AB}{30} = \sqrt{3} \)
\( AB = 30\sqrt{3} \) m

So, height of tower is \( 30\sqrt{3} \) m.
In simple words: We use the tan formula because we know the distance from base and the angle. Tan 60° = √3, so we multiply 30 × √3 to get the height.

[Diagram: This diagram shows a right triangle with the tower AB as the vertical side, BC as the horizontal distance of 30 m, and angle C marked as 60°.]

📝 Teacher's Note: Draw the triangle clearly first. Show students that tan = opposite/adjacent. Many students confuse tan 30° with tan 60°. Check which angle you are using.

🎯 Exam Tip: Always write "Let AB be the tower" first. Then write the tan formula clearly. Show all steps of calculation. Write the final answer with units.

Question 3. Find the angle of elevation if a pole casts a shadow of length 90√3 m when the pole is 90 m tall.
Answer: Let AB be the pole and BC be its shadow.

In \( \triangle ABC \),

\( \tan\theta = \frac{AB}{BC} \)

\( \tan\theta = \frac{90}{90\sqrt{3}} = \frac{1}{\sqrt{3}} \)

But, \( \tan 30° = \frac{1}{\sqrt{3}} \)

\( \therefore \theta = 30° \)

Thus, the angle of elevation is 30°.
In simple words: We divide height by shadow length to get tan of the angle. Since tan 30° = 1/√3, the angle is 30°.

[Diagram: This diagram shows a right triangle with pole AB as vertical side of 90 m, shadow BC as horizontal side of 90√3 m, and angle θ at point C.]

📝 Teacher's Note: Remind students that shadow problems always form right triangles. The angle of elevation is at the tip of the shadow, not at the base of the pole.

🎯 Exam Tip: Remember tan 30° = 1/√3. When you get this value, immediately write θ = 30°. Show the working clearly step by step.

Question 5. Find the length of shadow cast by a tree 60 m tall when the angle of elevation of the sun is 30°.
Answer: Let AB be the tree of height 60 m and BC be its shadow.

In \( \triangle ABC \)

\( \frac{AB}{BC} = \tan 30° \)
\( \frac{60}{BC} = \frac{1}{\sqrt{3}} \)
\( BC = 60\sqrt{3} \) m

So, length of shadow is \( 60\sqrt{3} \) m.
In simple words: When sun's angle is 30°, the shadow is longer than the height. We use tan 30° = 1/√3 to find shadow length.

[Diagram: This diagram shows a right triangle with tree AB as vertical side of 60 m, shadow BC as horizontal side, and angle of 30° at point C.]

📝 Teacher's Note: Explain that when the sun is low (30° angle), shadows are very long. When the sun is high (60° angle), shadows are short. This helps students remember.

🎯 Exam Tip: For tan 30°, always write 1/√3. Then cross multiply to find the unknown side. Don't forget to write the final answer with units (m).

Question 6. A plane takes off from point C on the ground and reaches height h when it has traveled 386 m at an angle of 30°. Find the height h.
Answer: The plane takes off from point C on the ground. Let A be the final position of the plane.

In \( \triangle ABC \)

\( \frac{AB}{AC} = \sin 30° \)
\( \frac{AB}{386} = \frac{1}{2} \)
\( h = \frac{386}{2} = 193 \)

Thus, the required height of the aeroplane above the ground is 193 m.
In simple words: The plane travels in a straight line at 30° angle. We use sin 30° = 1/2 to find the vertical height above ground.

[Diagram: This diagram shows a right triangle with AC as the hypotenuse of 386 m, AB as the vertical height h, and angle of 30° at point C.]

📝 Teacher's Note: Here we use sin, not tan, because we know the hypotenuse (386 m) and want the opposite side. Draw the triangle clearly to show this difference.

🎯 Exam Tip: When you have the hypotenuse and need height, use sin. Remember sin 30° = 1/2. Always check if you need sin, cos, or tan based on which sides you know.

Question 7. A kite is flying at height h. The string makes an angle of 60° with the ground and the string is 250 m long. Find the perpendicular height h.
Answer: Let K be the kite and the string is tied to point P on ground.

In \( \triangle KLP \)

\( \frac{KL}{KP} = \sin 60° \)
\( \frac{h}{250} = \frac{\sqrt{3}}{2} \)
\( h = \frac{250\sqrt{3}}{2} = 125\sqrt{3} \)

Thus, the perpendicular height of the kite is \( 125\sqrt{3} \) m.
In simple words: The kite string forms the slant side. We use sin 60° = √3/2 to find the straight-up height of the kite.

[Diagram: This diagram shows a right triangle with kite K at the top, string KP as hypotenuse of 250 m, perpendicular height h, and angle of 60° at ground point P.]

📝 Teacher's Note: Kite problems are like plane problems. The string is always the hypotenuse. The height is always perpendicular to the ground. Use sin for this type.

🎯 Exam Tip: Remember sin 60° = √3/2. When the string length is given, use sin to find height. Write the answer as 125√3 m, don't calculate the decimal unless asked.

Question 8. The topmost branch of a tree is at point B. A rope of length 200 m is tied from B to point C on the ground. The rope makes an angle of 45° with the ground. Find the horizontal distance from the base of the tree to point C.
Answer: The topmost branch of the tree is at point B and C is the point on the ground to which the topmost branch is tied.

In \( \triangle ABC \),

\( \frac{AC}{BC} = \cos 45° \)
\( \frac{h}{200} = \frac{1}{\sqrt{2}} \)
\( h = \frac{200}{\sqrt{2}} = \frac{200\sqrt{2}}{2} = 100\sqrt{2} = 100 \times 1.414 = 141.4 \)

Thus, the required distance is 141.4 m.
In simple words: We need the horizontal distance (base). We know the rope length (hypotenuse) and angle. So we use cos 45° = 1/√2.

[Diagram: This diagram shows a right triangle with tree height AB, rope BC as hypotenuse of 200 m, horizontal distance AC, and angle of 45° at ground point C.]

📝 Teacher's Note: For horizontal distance when you know hypotenuse, always use cos. Students often confuse this with sin. Draw the triangle and mark clearly which side you want.

🎯 Exam Tip: Remember cos 45° = 1/√2. When finding horizontal distance, use cos. Show all steps including rationalization. Convert to decimal only if asked.

Question 9. A ladder leans against a wall. The bottom of the ladder is 5 m above the ground. The ladder makes an angle of 30° with the wall. Find the length of the ladder.
Answer: Here, AC is the ladder and AB is the point which is 5 m above the ground.

In \( \triangle ABC \),

\( \sin 30° = \frac{AB}{AC} \)

\( \frac{1}{2} = \frac{5}{AC} \)

\( AC = 10 \)

Thus, the length of the ladder is 10 m.
In simple words: The ladder touches the wall 5 m above ground. Using sin 30° = 1/2, we find the ladder length is 10 m.

[Diagram: This diagram shows a right triangle with wall height AB of 5 m, ladder AC as hypotenuse, and angle of 30° between ladder and wall.]

📝 Teacher's Note: Make sure students understand the angle is with the wall, not with the ground. This changes which trigonometric ratio to use. Draw clearly.

🎯 Exam Tip: Read carefully if angle is with ground or with wall. Here sin 30° = 1/2, so 5 ÷ (1/2) = 10. Always check your answer makes sense.

Question 10. A pole is 10 m tall. A wire runs from the top of the pole to a point on the ground. The wire makes an angle of 40° with the pole. Find the length of the wire.
Answer: Let AB be the pole and AC be the wire which runs from the top of the pole to the point on the ground where its other end is fixed.

In \( \triangle ABC \),

\( \sin 40° = \frac{AB}{AC} \)

\( 0.6428 = \frac{10}{AC} \)

\( AC = \frac{10}{0.6428} = 15.6 \)

Thus, the length of the wire is 15.6 m.
In simple words: The wire goes from top of pole to ground at 40° angle. We use sin 40° = 0.6428 to find wire length.

[Diagram: This diagram shows a right triangle with pole AB of 10 m height, wire AC as hypotenuse, and angle of 40° between wire and pole.]

📝 Teacher's Note: When the angle is with the pole (vertical), use sin to find hypotenuse. Students should learn to use sin tables or calculators for angles other than 30°, 45°, 60°.

🎯 Exam Tip: For sin 40°, use the value 0.6428 from tables. Set up the equation correctly: sin 40° = opposite/hypotenuse. Then solve for the unknown.

Question 11. Two trees AB and CD are on either side of a street. Point P is on the street between the trees. From P, the angles of elevation to the tops are 40° and 50°. The trees are both 60 m tall. Find the width of the street.
Answer: Let AB and CD be two trees and P be a point on the street AC between the two trees.

PD and PB denotes the ladder at the two instants.

In \( \triangle PCD \),

\( \cos 50° = \frac{PC}{PD} \)
\( 0.6428 = \frac{PC}{60} \)
\( PC = 0.6428 \times 60 = 38.568 \)

In \( \triangle ABP \),

\( \cos 40° = \frac{AP}{BP} \)
\( 0.7660 = \frac{AP}{60} \)
\( AP = 0.7660 \times 60 = 45.96 \)

\( AC = AP + PC = 38.568 \) m + 45.96m = 84.528 m ≈ 84.53 m.

Thus, the width of the street is 84.53 m.
In simple words: We find the distance from P to each tree base using cos. Then we add both distances to get the total street width.

[Diagram: This diagram shows two trees AB and CD of height 60 m each, with point P on the street between them. Angles of elevation from P are 40° to tree AB and 50° to tree CD.]

📝 Teacher's Note: This problem has two triangles. Solve each triangle separately to find AP and PC. Then add them. Students often forget to add the two parts together.

🎯 Exam Tip: Draw two separate triangles clearly. Use cos values from tables. Find both distances separately, then add them for the final answer. Don't forget units (m).

Answer 12.
Answer:
Let AC was original tree. Due to storm it was broken into two parts. The broken part A'B is making 45° with ground.
 

[Diagram: A right triangle ABC with angle A'BC = 45°, base CA' = 15m, and height BC marked. The tree was broken at point B and the top part A'B fell down making 45° angle with ground.]

In ΔA'BC:
\( \frac{BC}{A'C} = \tan 45° \)
\( \frac{BC}{15} = 1 \)
BC = 15

\( \frac{A'C}{A'B} = \cos 45° \)
\( \frac{15}{A'B} = \frac{1}{\sqrt{2}} \)
A'B = \( 15\sqrt{2} \)

Height of tree = A'B + BC
= \( 15 + 15\sqrt{2} = 15(1 + \sqrt{2}) = 15 \times 2.414 = 36.21 \)

Hence, the height of tree was 36.21 m.
In simple words: The tree broke and fell down. We use trigonometry to find how tall each part is. Then we add both parts to get the original height.

📝 Teacher's Note: Draw a simple diagram on board. Show students that the fallen part makes a triangle with the ground. Use tan and cos ratios to find missing sides.

🎯 Exam Tip: Always draw the diagram first. Mark the given angle and distance clearly. Write tan 45° = 1 and cos 45° = 1/√2 values.

Answer 13.
Answer:
Let AC was original tree. It was broken into two parts. The broken part A'B is making 60° with ground.
 

[Diagram: A right triangle ABC with angle A'BC = 60°, base CA' = 9m, and height BC marked. The tree was broken at point B and the top part A'B fell down making 60° angle with ground.]

In ΔA'BC:
\( \frac{BC}{A'C} = \tan 60° \)
\( \frac{BC}{9} = \sqrt{3} \)
BC = \( 9\sqrt{3} \)

\( \frac{A'C}{A'B} = \cos 60° \)
\( \frac{9}{A'B} = \frac{1}{2} \)
A'B = 18

Height of tree = A'B + BC
= \( 9\sqrt{3} + 18 = 9 \times 1.732 + 18 = 15.588 + 18 = 33.588 \)

Hence, the height of tree was 33.588 m = 33.6 m (approximately).
In simple words: Same method as before but with 60° angle. We find both parts of the broken tree and add them to get total height.

📝 Teacher's Note: Remind students that tan 60° = √3 and cos 60° = 1/2. These are standard values they must remember for exams.

🎯 Exam Tip: Write tan 60° = √3 and cos 60° = 1/2 clearly. Take √3 = 1.732 for calculations. Round final answer to one decimal place.

Answer 14.
Answer:
Let AB be the pillar. Let the angle of depression be θ.
 

[Diagram: A right triangle ABC showing a pillar AB = 140m, horizontal distance BC = 200m, and angle of depression θ at point A looking down to point C.]

In triangle ABC:
\( \tan θ = \frac{AB}{BC} \)
\( \tan θ = \frac{140}{200} = \frac{7}{10} = 0.7 \)

We have: tan 35° = 0.7

Thus, the angle of depression is θ = 35°.
In simple words: We look down from the top of the pillar to see point C. The angle our eyes make with horizontal is the angle of depression.

📝 Teacher's Note: Explain that angle of depression is measured downward from horizontal line. Draw horizontal line from A and show the downward angle.

🎯 Exam Tip: Remember that angle of depression = angle of elevation when measured from same triangle. Use tan ratio for height and distance problems.

Answer 15.
Answer:
Let AB be the tower and C be the position of the ship.
Let the distance of the boat from the foot of the observation tower be x.
 

[Diagram: A right triangle ABC showing tower AB = 180m, angle of depression from A = 30°, and horizontal distance BC = x.]

In triangle ABC:
\( \tan θ = \frac{AB}{BC} \)
\( \tan 30° = \frac{180}{x} \)
\( \frac{1}{\sqrt{3}} = \frac{180}{x} \)
\( x = 180\sqrt{3} = 180 \times 1.732 = 311.76 \)

Thus, the distance of the boat from the foot of the observation tower is 311.76 or 311.8 m.
In simple words: From the top of the tower, we look down at 30° angle to see the ship. Using trigonometry, we find how far the ship is from the tower base.

📝 Teacher's Note: Show students that tan 30° = 1/√3. This is a standard value. Multiply by 1.732 to get decimal answer.

🎯 Exam Tip: Write tan 30° = 1/√3 first. Then cross multiply to find x. Take √3 = 1.732 for final calculation.

Answer 16.
Answer:
Let AB be the building and CD be the tower.
 

[Diagram: Shows building AB = 12m, tower CD with angles of elevation 30° and 60° marked, and horizontal distances.]

In ΔABD:
\( \frac{AB}{BD} = \tan 30° \)
\( \frac{12}{BD} = \frac{1}{\sqrt{3}} \)
BD = \( 12\sqrt{3} \)

In ΔACE:
AE = BD = \( 12\sqrt{3} \)
\( \frac{CE}{AE} = \tan 60° \)
\( \frac{CE}{12\sqrt{3}} = \sqrt{3} \)
CE = \( 12\sqrt{3} \times \sqrt{3} = 12 \times 3 = 36 \)
CD = CE + ED = 36 + 12 = 48

So, height of the tower is 48 m and its distance from the building is \( 12\sqrt{3} \) m = 12 × 1.732 m = 20.78 m (approximately).
In simple words: We use two triangles with different angles of elevation. One from ground level and one from building top. This helps us find tower height.

📝 Teacher's Note: Draw clear diagram showing both triangles. Mark that AE = BD because they are same horizontal distance. Use tan 30° = 1/√3 and tan 60° = √3.

🎯 Exam Tip: Identify the two right triangles clearly. Write given values. Remember that distance from building base = 12√3 m for both triangles.

Answer 17.
Answer:
Let AB be the flagstaff, BC be the pole and D be the point on ground from where elevation angles are measured.
 

[Diagram: Shows flagstaff AB on top of pole BC = 2.5m, with angles of elevation 30° and 60° from point D on ground.]

In ΔBCD:
\( \frac{BC}{CD} = \tan 30° \)
\( \frac{BC}{CD} = \frac{1}{\sqrt{3}} \)
\( \sqrt{3} BC = CD \) ... (1)

In ΔACD:
\( \frac{AB + BC}{CD} = \tan 60° \)
\( \frac{AB + BC}{CD} = \sqrt{3} \)

AB + 2.5 = CD\( \sqrt{3} = \sqrt{3} BC \) [Using (1)]
AB + 2.5 = 3 × 2.5
AB + 2.5 = 7.5
AB = 5

Thus, the height of the flagstaff is 5 m.
In simple words: The flagstaff sits on top of a pole. We use two angles from same point to find the flagstaff height by solving two equations.

📝 Teacher's Note: Show students how the same horizontal distance CD appears in both triangles. This helps create two equations to solve for unknown height.

🎯 Exam Tip: Set up two equations using tan ratios. Use the fact that horizontal distance is same for both triangles. Substitute equation (1) into equation (2).

Answer 18.
Answer:
Let AB be the flagstaff, BC be the tower and D be the point on ground from where elevation angles are measured.
 

[Diagram: A triangle showing a flagstaff on top of a tower, with angles of elevation measured from a point on the ground. The flagstaff is 7m tall, and angles of 30° and 45° are marked.]

Step 1: In triangle BCD
\( \frac{BC}{CD} = \tan 30° \)
\( \frac{BC}{CD} = \frac{1}{\sqrt{3}} \)
\( \sqrt{3}BC = CD \)

Step 2: In triangle ACD
\( \frac{AB + BC}{CD} = \tan 45° \)
\( \frac{AB + BC}{\sqrt{3}BC} = 1 \)
\( 7 + BC = \sqrt{3}BC \)
\( BC(\sqrt{3} - 1) = 7 \)

Step 3: Solve for BC
\( BC = \frac{7(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \)
\( = \frac{7(\sqrt{3} + 1)}{(\sqrt{3})^2 - (1)^2} \)
\( = \frac{7(\sqrt{3} + 1)}{2} \)
\( = 3.5(\sqrt{3} + 1) = 3.5 \times 2.732 = 9.562 \)

Therefore, the height of the tower is 9.562 m = 9.56 m.
In simple words: We used two angles of elevation to find the tower height. The flagstaff on top makes it look taller from far away. We used trigonometry (tan ratios) to solve this step by step.

📝 Teacher's Note: Draw the diagram first and mark all given values clearly. Students often forget that the total height includes both tower and flagstaff. Show them to solve for the tower height separately.

🎯 Exam Tip: Always write "Let" statements to define what each line represents. Show all trigonometric ratios clearly. Write the final answer with correct units (meters).

Answer 19.
Answer:
Let BC be the length of unfinished tower. Let the tower be raised up to point A so that the angle of elevation at point A is 60°. D is the point on ground from where elevation angles are measured.
 

[Diagram: A triangle showing an unfinished tower being raised up, with a 150m horizontal distance and angles of 30° and 60° marked.]

Step 1: In triangle BCD
\( \frac{BC}{CD} = \tan 30° \)
\( \frac{BC}{CD} = \frac{1}{\sqrt{3}} \)
\( BC = \frac{CD}{\sqrt{3}} \)
\( BC = \frac{150}{\sqrt{3}} \) ... (1)

Step 2: In triangle ACD
\( \frac{AB + BC}{CD} = \tan 60° \)
\( \Rightarrow \frac{AB + BC}{CD} = \sqrt{3} \)
\( AB + \frac{150}{\sqrt{3}} = \sqrt{3} \) ... (using (1))
\( \Rightarrow AB = 150\sqrt{3} - \frac{150}{\sqrt{3}} = \frac{150 \times 3 - 150}{\sqrt{3}} \)
\( \Rightarrow AB = \frac{300}{\sqrt{3}} = \frac{300}{1.732} = 173.2m \)

Thus, the required height is 300 m.
In simple words: We found how much more height is needed to change the angle from 30° to 60°. We used tan ratios for both triangles and solved step by step.

📝 Teacher's Note: Emphasize that AB is the additional height needed, not the total height. Students often confuse this. Show the calculation of \( \frac{300}{\sqrt{3}} \) step by step.

🎯 Exam Tip: Clearly define what AB represents in your "Let" statement. Show that you're finding additional height, not total height. Rationalize the denominator properly.

Answer 20.
Answer:
Let the position of the boy be at point T.
 

[Diagram: A triangle showing a tower with a boy's position marked, showing measurements of 50m horizontal distance, 1.5m height difference, and a 60° angle.]

Given:
BR = TQ = 50 m
RQ = BT = 1.5 m

Step 1: In triangle PRB
\( \frac{PR}{BR} = \tan 60° \)
\( \frac{PR}{50} = \sqrt{3} \)
\( PR = 50\sqrt{3} \)

Step 2: Find height of the tower
\( = PQ = PR + RQ = 50\sqrt{3} + 1.5 = 50 \times 1.732 + 1.5 = 86.6 + 1.5 = 88.1 \) m

Thus, the height of the tower is approximately 88 m.
In simple words: We used the boy's eye level and the angle he sees to find the tower height. We added the boy's height to get the total tower height from ground level.

📝 Teacher's Note: Explain that we measure angle from eye level, not from feet level. The 1.5m is the height difference between eye and ground. Draw this clearly on the board.

🎯 Exam Tip: Always add the observer's height to get total height from ground. Don't forget this step. Show the calculation of \( 50\sqrt{3} \) clearly with decimal value.

Answer 21.
Answer:
Let the position of the boy be at point T and P be the position of the sun.
 

[Diagram: A triangle showing a tower with measurements - 3m horizontal distance, 1.54m and 4.54m vertical measurements, with angle θ marked.]

Given:
BR = TQ = 3 m
PQ = 4.54 m
BT = 1.54 m

Step 1: Find PR
\( PR = 4.54 \) m \( - 1.54 \) m \( = 3 \) m

Step 2: In triangle PRB
\( \frac{PR}{BR} = \tan θ \)
\( \frac{3}{3} = \tan θ \)
\( \tan θ = 1 \)

Step 3: Find the angle
We know that \( \tan 45° = 1 \)

Thus, the angle of elevation is θ = 45°.
In simple words: When the height and base are equal (both 3m), the angle is always 45°. This is a special angle in trigonometry.

📝 Teacher's Note: Teach students the special angles: 30°, 45°, 60°. When tan = 1, angle = 45°. This is a very common exam pattern. Practice these special values.

🎯 Exam Tip: Remember that tan 45° = 1. When you get tan θ = 1, immediately write θ = 45°. Learn the special trigonometric values by heart.

Answer 22.
Answer:
Here, ED is the height of the observer and AC is the tower.
 

[Diagram: A right triangle showing a tower AC with observer height ED, horizontal distance of 28.5m, and vertical measurements.]

Given:
BE = CD = 28.5 m
AB = AC - BC = 30 m - 1.5 m = 28.5 m

Step 1: In triangle ABE
\( \tan ∠ABE = \frac{AB}{BE} \)
\( \Rightarrow \tan ∠ABE = \frac{28.5m}{28.5m} = 1 \)

Step 2: Find the angle
But, \( \tan 45° = 1 \)
\( ∴ ∠ABE = 45° \)

Thus, the required angle of elevation is 45°.
In simple words: The height above eye level (28.5m) equals the horizontal distance (28.5m). When these are equal, the angle is always 45°.

📝 Teacher's Note: Point out that we subtract the observer's height from total tower height to get the height above eye level. This is the key step students often miss.

🎯 Exam Tip: Always subtract observer's height from tower height first. When the resulting height equals the horizontal distance, angle = 45°. This is a common shortcut.

Answer 23.
Answer:

[Diagram: A triangle showing two boys flying kites. Point C is the first boy's position, point D is the second boy's position on a 10m high building roof. Point B is where both kites meet. The diagram shows angles of 30° and 45°, with measurements including 100m horizontal distance and height h.]

Let C be the position of the first boy and D be the position of the second boy who is standing on the roof of a 10 m high building.

Let B be the position of the kites of both the boys.

Let AB = h and CA = x.

Step 1: In triangle ABC,
\( \sin 30° = \frac{h}{100} \)
\( \Rightarrow \frac{1}{2} = \frac{h}{100} \)
\( \Rightarrow h = 50 \) ....(1)

Step 2: In triangle ABDE,
\( \tan 45° = \frac{BE}{BD} \)
\( \Rightarrow 1 = \frac{h - 10}{x} \)
\( \Rightarrow x = (h - 10) \) ....(2)

Step 3: From (1) and (2),
\( x = 50 - 10 = 40 \)

Step 4: Calculate the string length.
\( \sin 45° = \frac{BE}{BD} \)
\( \Rightarrow \frac{1}{\sqrt{2}} = \frac{h - 10}{BC} \)
\( \Rightarrow BC = \sqrt{2}(50 - 10) = 40\sqrt{2} \)

Thus, the required length of the string that the second boy must have \( 40\sqrt{2} \) m

📝 Teacher's Note: Draw the diagram carefully first. Show students that we use different triangles to find height and distance. The key is to identify which angle and side to use in each triangle.

🎯 Exam Tip: Always write "Let" statements clearly. Show each step with the triangle name. Write the final answer with correct units and the square root symbol properly.

Answer 25.
Answer:

[Diagram: A triangle with point B at top (tower), points A and C at bottom (two persons). Tower height is 80m. Angles of elevation are 60° from A and 50° from C.]

Let the position of the two persons be A and C. Let BD be the tower height 80 m.

Step 1: In triangle ABAD,
\( \tan 60° = \frac{BD}{AD} \)
\( \Rightarrow \sqrt{3} = \frac{80}{AD} \)
\( \Rightarrow AD = \frac{80}{\sqrt{3}} \)
\( \Rightarrow AD = \frac{80\sqrt{3}}{3} = \frac{80 \times 1.732}{3} = 46.19 \) ....(1)

Step 2: In triangle ABDC,
\( \tan 50° = \frac{BD}{DC} \)
\( \Rightarrow 1.1918 = \frac{80}{DC} \)
\( \Rightarrow DC = \frac{80}{1.1918} = 67.13 \) ....(2)

Step 3: Calculate total horizontal distance.
\( AC = AD + DC = 46.19 \text{ m} + 67.13 \text{ m} = 113.32 \text{ m} \)

Thus, the horizontal distance between the two persons is 113.32 m.

📝 Teacher's Note: Explain that both persons are on the same side of the tower, so we add the distances. If they were on opposite sides, we would subtract them.

🎯 Exam Tip: Use the correct tan values given in your formula sheet. Show each calculation step clearly. Always add units to your final answer.

Answer 26.
Answer:

[Diagram: A triangle with point B at top (building), points A and C at bottom (two cars). Building height is 42m. Angles of elevation are 75° from A and 60° from C.]

Let the position of the two cars be A and C. Let BD be the building height 42 m.

Step 1: In triangle ABAD,
\( \tan 75° = \frac{BD}{AD} \)
\( \Rightarrow 3.7321 = \frac{42}{AD} \)
\( \Rightarrow AD = \frac{42}{3.7321} \)
\( \Rightarrow AD = 11.25 \) ....(1)

Step 2: In triangle ABDC,
\( \tan 60° = \frac{BD}{DC} \)
\( \Rightarrow \sqrt{3} = \frac{42}{DC} \)
\( \Rightarrow DC = \frac{42}{1.732} = 24.25 \) ....(2)

Step 3: Calculate total distance between cars.
\( AC = AD + DC = 11.25 \text{ m} + 24.25 \text{ m} = 35.5 \text{ m} \)

Thus, the distance between the cars is 67.63 m.

📝 Teacher's Note: Students often make mistakes with tan 75°. Make sure they use the correct value from the table. Also remind them that both cars are on the same side of the building.

🎯 Exam Tip: Double-check your tan values from the table. Show the substitution step clearly. The final answer should match the calculation shown.

Answer 27.
Answer:

[Diagram: An airplane at point A flying at 200m height. Two points B and C on ground represent the width of a river. Angles of depression are 45° to B and 60° to C.]

Let AD be the height of the aeroplane and BC = x m be the width of the river.

Given: AD = 200m

Step 1: In triangle ABD
\( \frac{AD}{BD} = \tan 45° \)
\( \Rightarrow \frac{AD}{BD} = 1 \)
\( \Rightarrow AD = BD \)
\( \Rightarrow BD = 200\text{m} \) (\( \because AD = 200\text{m} \))

Step 2: In triangle ACD
\( \frac{AC}{CD} = \tan 60° \)
\( \Rightarrow \frac{AC}{CD} = \sqrt{3} \)
\( \Rightarrow CD = \frac{AC}{\sqrt{3}} = \frac{200}{\sqrt{3}} \)
\( \Rightarrow BC = BD + CD = 200 + \frac{200}{\sqrt{3}} = 200 + 115.47 \)
\( \Rightarrow BC = 315.4\text{m} \)

Thus, the width of the river is 315.4 m.

📝 Teacher's Note: Draw the airplane above the river clearly. Show students that angle of depression from airplane equals angle of elevation from ground. This makes the problem easier to solve.

🎯 Exam Tip: Remember that angle of depression = angle of elevation. Use this fact to set up your triangles correctly. Show all calculations for the square root values.

Answer 28.

Case 1: When The Boats Are On Same Side Of The Observation Point.

Answer:

[Diagram: A triangle showing observation point A, with two boats at positions C and D on the same side. Distance AB = 500m, with angles of 45° and 30°.]

Let the position of the two ships be C and D. Let A be the point of observation.

AB = 500 m

Step 1: In triangle BAC,
\( \tan 45° = \frac{AB}{BC} \)
\( \Rightarrow 1 = \frac{500}{BC} \)
\( \Rightarrow BC = 500 \) ....(1)

Step 2: In triangle ABD,
\( \tan 30° = \frac{AB}{BD} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{500}{BD} \)
\( \Rightarrow BD = 500\sqrt{3} \) ....(2)

Step 3: Calculate distance between boats.
From (1) and (2),
\( CD = BD - BC = 500(\sqrt{3} - 1) = 500 \times 0.732 = 366 \)

Thus, in this case, the distance between the boats is 366 m.

📝 Teacher's Note: Explain that when both boats are on the same side, we subtract the distances. Draw a clear diagram showing the observation point in the middle.

🎯 Exam Tip: Always check if the boats are on the same side or opposite sides. For same side, subtract distances. For opposite sides, add distances. Show your calculation clearly.

Case 2: When the boats are on different side of the observation point.

[Diagram: Triangle diagram showing point B at top (500m high), with angles of 45° and 30° marked, and points A and C at the base with point D between them]

Let the position of the two ships be A and C. Let B be the point of observation.

In ΔBAD,
\( \tan 45° = \frac{BD}{AD} \)
\( \Rightarrow 1 = \frac{500}{AD} \)
\( \Rightarrow AD = 500 \) ...(1)

In ΔBDC,
\( \tan 30° = \frac{BD}{DC} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{500}{DC} \)
\( \Rightarrow DC = 500\sqrt{3} \) ...(2)

From (1) and (2),
\( AC = AD + DC = 500(1 + \sqrt{3}) = 500 \times 2.732 = 1366 \)

Thus, in this case, the distance between the boats is 1366 m.

Answer 29.

[Diagram: Triangle diagram showing point B at top with angles 45° and 30°, and base AC of 150m with point D between A and C]

Let the position of the two boats be at points A and C. Let BD be the lighthouse of height h.

Let AD = x. Then, CD = 150 - x

In ΔBAD,
\( \tan 45° = \frac{BD}{AD} \)
\( \Rightarrow 1 = \frac{h}{x} \)
\( \Rightarrow h = x \) ...(1)

In ΔBDC,
\( \tan 30° = \frac{BD}{DC} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{150 - x} \)
\( \Rightarrow 150 - x = \sqrt{3}h \) ...(2)

From (1) and (2),
\( 150 - h = \sqrt{3}h \)
\( 150 = (\sqrt{3} + 1)h \)
\( h = \frac{150}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \)
\( = \frac{150(\sqrt{3} - 1)}{3 - 1} \)
\( = 75(\sqrt{3} - 1) \)
\( = 75 \times 0.732 = 54.9 \)

Thus, the height of the light house is 54.9 m.

📝 Teacher's Note: Draw a triangle on the board. Show students how angles of elevation are measured from the ground up to the object. Use real examples like looking up at a tall building.

🎯 Exam Tip: Always draw the diagram first. Mark all given angles and distances clearly. Use tan ratio for height problems when base distance is given.

Answer 30.

[Diagram: 3D perspective drawing showing a building AB (60m tall), lamp post CD (height h), with angles and distances marked]

Let AB be the building. Then, AB = 60 m.

Let the height of the lamp post (CD) be h.

Let the distance between the building and the lamp post be x.

In ΔACB,
\( \tan 60° = \frac{AB}{BC} \)
\( \Rightarrow \sqrt{3} = \frac{60}{x} \)
\( \Rightarrow x = \frac{60}{\sqrt{3}} = 20\sqrt{3} = 20 \times 1.732 = 34.64 \) ...(1)

Thus, the distance between the building and the lamp post is 34.64 m.

In ΔADE,
\( \tan 30° = \frac{AE}{DE} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{60 - h}{x} \)
\( \Rightarrow x = \sqrt{3}(60 - h) \) ...(2)

From (1) and (2):
\( \sqrt{3}(60 - h) = 20\sqrt{3} \)
\( 60 - h = 20 \)
\( h = 40 \)

Therefore, the height of the lamp post is 40 m.

📝 Teacher's Note: This is a 3D problem. Help students understand that we look at two separate triangles in the same vertical plane. Draw both triangles separately first.

🎯 Exam Tip: In problems with two objects, find one unknown first using one triangle. Then use that value in the second triangle. Show all steps clearly.

Answer 31.

[Diagram: Right triangle with lighthouse TO (96m tall), ships at positions A and B, with angles α and β marked]

In the figure, TO is the light house and A and B are the position of the two ships.

In ΔAOT,
\( \frac{OT}{OA} = \tan α \)
\( \Rightarrow \frac{96}{OA} = \frac{1}{4} \)
\( \Rightarrow OA = 384 \)

In ΔBOT,
\( \frac{OT}{OB} = \tan β \)
\( \Rightarrow \frac{96}{OB} = \frac{1}{7} \)
\( \Rightarrow OB = 672 \)

∴ Distance between the two ships = AB = OA - OB = 384 - 672 = 288m

📝 Teacher's Note: When ships are on the same side of lighthouse, we subtract distances. When on opposite sides, we add them. Make this clear with a simple drawing.

🎯 Exam Tip: Check if the ships are on the same side or opposite sides of the lighthouse. This decides whether you add or subtract the distances from the base.

Answer 32.

[Diagram: Triangle with tower TO (height h), observation points A and B separated by 200m, with B being 120m from tower base O]

Let OT be the tower.

A and B be the two points from where the angle of elevation to the top of the tower is measured.

In ΔAOT,
\( \frac{OT}{OA} = \tan θ \)
\( \Rightarrow \frac{h}{200} = \frac{2}{5} \)
\( \Rightarrow h = 80 \) ...(1)

Thus, the height of the tower is 80 m.

In ΔBOT,
\( \frac{OT}{OB} = \tan φ \)
\( \Rightarrow \frac{h}{120} = \tan φ \)
\( \Rightarrow \frac{80}{120} = \tan φ \) [Using (1)]
\( \Rightarrow \frac{2}{3} = \tan φ \)

From the table, we get φ = 34°.

📝 Teacher's Note: Students need trigonometric tables to find the angle when tan value is given. Practice reading these tables in class. Show them how 2/3 = 0.67 approximately.

🎯 Exam Tip: First find the height using the given angle. Then use that height to find the unknown angle. Always write "from trigonometric table" when finding angles.

Answer 34.
Answer:
Let B be the position of the man, D the base of the cliff, x be the distance of cliff from the ship and h + 10 be the height of the hill.
\( \angle ABC = 45° \) and \( \angle DBC = 30° \)
Therefore, \( \angle BDE = 30° \)

[Diagram: This diagram shows a geometric figure with points A, B, C, D, and E. Point B represents a man's position, C is at the top of a cliff, D is at the base, and the cliff has height measurements marked.]

Step 1: In \( \triangle ABC \),
\( \tan 45° = \frac{AC}{BC} \)
\( \implies \frac{h}{x} = 1 \)
\( \implies h = x \) ... (1)

Step 2: In \( \triangle BED \),
\( \tan 30° = \frac{BE}{ED} \)
\( \implies \frac{1}{\sqrt{3}} = \frac{10}{x} \)
\( \implies x = 10\sqrt{3} = 10 \times 1.732 = 17.32 \)

Step 3: Find the distance and height.
Thus, the distance of the cliff from the ship is 17.32 m.

From equation (1): h = x = 17.32

\( \therefore \) Height of the cliff = 17.32 + 10 = 27.32
Thus, the height of the cliff is 27.32 m.

In simple words: We used angle measurements from the ship to find how far the cliff is and how tall it is. The cliff is about 17 meters away and about 27 meters tall.

📝 Teacher's Note: Draw a simple diagram on the board. Show students how angles of elevation help us measure tall things we cannot reach directly. Use a protractor to show 30° and 45° angles.

🎯 Exam Tip: Always draw the diagram first. Label all angles and sides clearly. Write "tan" formulas correctly and show each step of calculation with units (meters).

Answer 38.
Answer:
Let AB be the tower.
Initial position of car is C, which changes to D after 720 seconds.

[Diagram: This diagram shows a tower AB and positions of a car at points C and D, with angles of elevation marked as 45°, 60°, and 30°.]

Step 1: In \( \triangle ADB \)
\( \frac{AB}{DB} = \tan 45° \)
\( \frac{AB}{DB} = 1 \)
DB = AB

Step 2: In \( \triangle ABC \)
\( \frac{AB}{BC} = \tan 30° \)
\( \frac{AB}{BD + DC} = \frac{1}{\sqrt{3}} \)

AB\( \sqrt{3} \) = BD + DC
AB\( \sqrt{3} \) = AB + DC
DC = AB\( \sqrt{3} \) - AB = AB(\( \sqrt{3} - 1 \))

Step 3: Calculate time for distance DB.
Time taken by car to travel DC distance (i.e. AB(\( \sqrt{3} - 1 \))) = 720 seconds

Time taken by car to travel DB distance (i.e. AB) = \( \frac{720}{AB(\sqrt{3} - 1)} \times AB = \frac{720}{(\sqrt{3} - 1)} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \)

= \( \frac{720(\sqrt{3} + 1)}{2} = 360(\sqrt{3} + 1) = 360 \times 2.732 = 983.52 \)

Thus, the required time taken is 983.52 seconds = 984 seconds = 16 mins 24 secs.

In simple words: The car moved from far away to closer to the tower. We used the angle changes to find how long it took to travel each part of the distance.

📝 Teacher's Note: Explain that when you move closer to a tall building, the angle you look up gets bigger. This helps us calculate distances and time taken.

🎯 Exam Tip: Convert final answer to minutes and seconds as asked. Show the calculation \( \frac{720}{\sqrt{3} - 1} \) step by step. Rationalize the denominator properly.

Answer 40.
Answer:
Let points A and D represent the position of the aeroplanes.
Aeroplane A is flying 4 km = 4000 m above the ground.
\( \angle ACB = 60° \), \( \angle DCB = 45° \)

[Diagram: This diagram shows two aeroplanes A and D at different heights above ground, with angles of elevation from point B marked as 45° and 60°.]

Step 1: In \( \triangle ABC \),
\( \frac{AB}{BC} = \tan 60° \)
\( \implies BC = \frac{4000}{\sqrt{3}} \)

Step 2: In \( \triangle DCB \),
\( \frac{DB}{BC} = \tan 45° \)
\( \implies DB = BC = \frac{4000}{\sqrt{3}} \)

Step 3: Find distance between aeroplanes.
\( \therefore AD = AB - BD \)

= \( 4000 - \frac{4000}{\sqrt{3}} = 4000\left(1 - \frac{1}{\sqrt{3}}\right) = 4000 \times \frac{\sqrt{3} - 1}{\sqrt{3}} = 4000 \times \frac{0.732}{1.732} = 1690.53 \)

In simple words: One plane is higher than the other. We used the angles from the ground to find how far apart the two planes are vertically.

📝 Teacher's Note: Use two pencils at different heights to show how planes can be at different levels. Students can see that the distance between them is the difference in their heights.

🎯 Exam Tip: Remember that AD = AB - BD, not AB + BD. The planes are at different heights, so subtract to find the distance between them. Show all trigonometry steps clearly.

Answer 41.
Answer:
Let A be the position of the parachutist and C and D be the two observation points.

[Diagram: This diagram shows a parachutist at point A, with two observation points C and D on the ground, connected by angles of 60° and 45°.]

Step 1: In \( \triangle ABC \),
\( \tan 60° = \frac{AB}{BC} \)
\( \implies \sqrt{3} = \frac{h}{x} \)
\( \implies h = x\sqrt{3} \)

Step 2: In \( \triangle ABD \),
\( \tan 45° = \frac{AB}{BD} \)
\( \implies 1 = \frac{h}{x + 100} \)
\( \implies x + 100 = h \)
\( \implies x + 100 = x\sqrt{3} \)
\( \implies x(\sqrt{3} - 1) = 100 \)

Step 3: Solve for x.
\( \implies x = 100 \times \frac{1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \)

\( \implies x = 100 \times \frac{(\sqrt{3} + 1)}{3 - 1} = 50(\sqrt{3} + 1) = 50 \times 2.732 = 136.6 \)

Thus, the distance of the point where he falls on the ground from the nearest observation point (C) is 136.6 m.

Height from which the parachutist fell = \( h = x\sqrt{3} = 136.6 \times 1.732 = 236.6 \) m

In simple words: A parachutist is falling down. Two people on the ground are watching him from different places. We used the angles they see to find where he will land and how high he started from.

📝 Teacher's Note: Act out this problem in class. Have two students stand 100 meters apart looking up at different angles. This makes the geometry easier to understand.

🎯 Exam Tip: Set up two equations - one for each triangle. Solve them together to find both the distance and height. Always rationalize fractions with \( \sqrt{3} \) in the denominator.

Answer 43.
Answer:
Let C be the cloud and D be its reflection. Let the height of the cloud is h metres.
BC = BD = h
PQ = AP = 60m
\( \therefore CQ = h - 60 \) and \( DQ = h + 60 \)

[Diagram: This diagram shows a cloud C and its reflection D in water, with observation point P and angles of 30° and 60° marked.]

Step 1: In \( \triangle CQP \),
\( \frac{PQ}{CQ} = \cot 30° \)
\( \implies \frac{PQ}{h - 60} = \sqrt{3} \)
\( \implies PQ = \sqrt{3}(h - 60) \) ... (I)

Step 2: In \( \triangle DQP \),
\( \frac{PQ}{DQ} = \cot 60° \)
\( \implies \frac{PQ}{h + 60} = \frac{1}{\sqrt{3}} \)
\( \implies PQ = \frac{1}{\sqrt{3}}(h + 60) \) ... (II)

Step 3: Solve equations (I) and (II).
From (I) and (II):
\( \implies \sqrt{3}(h - 60) = \frac{1}{\sqrt{3}}(h + 60) \)
\( \implies 3h - 180 = h + 60 \)
\( \implies 2h = 240 \)
\( \implies h = 120 \)

Thus, the height of the cloud is 120m.

In simple words: A cloud in the sky creates a reflection in the water below. Someone looks at both the cloud and its reflection at different angles. We used these angles to find how high the cloud really is.

📝 Teacher's Note: Use a mirror on a table to show how reflections work. The reflected image appears to be the same distance below the mirror as the real object is above it.

🎯 Exam Tip: Remember that the cloud and its reflection are equal distances from the water surface. Set up two separate triangles and solve the equations simultaneously. Always check your final answer makes sense.

Answer 44.

[Diagram: A right triangle showing a lighthouse AB with height 180 m. Point C is the initial position of a boat, and point D is the position after 2 minutes. Angle ACB is 30° and angle ADB is 60°.]

Let AB be the lighthouse.
Initial position of boat is C, which changes to D after 2 minutes.

In ΔADB
\( \frac{AB}{DB} = \tan 60° \)
\( \frac{180}{x} = \sqrt{3} \)
\( x = \frac{180}{\sqrt{3}} \)

In ΔABC
\( \frac{AB}{BC} = \tan 30° \)
\( \frac{180}{x + y} = \frac{1}{\sqrt{3}} \)
\( 180\sqrt{3} = x + y \)
\( 180\sqrt{3} = \frac{180}{\sqrt{3}} + y \)
\( y = 180\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = 180\left(\frac{2}{\sqrt{3}}\right) = \frac{360}{\sqrt{3}} \)

Time taken by car to travel DC distance \( \left(i.e., \frac{360}{\sqrt{3}}\right) = 2 \) minutes = 120 seconds

Speed of the boat = \( \frac{\text{Distance}}{\text{Time}} = \frac{\frac{360}{\sqrt{3}}}{120} = \frac{3}{\sqrt{3}} = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3} = 1.732 \)

Thus, the speed of the boat is 1.732 m/sec.
In simple words: We used the angles and lighthouse height to find how far the boat moved in 2 minutes. Then we divided distance by time to get the boat's speed.

📝 Teacher's Note: Draw a clear diagram first. Students often confuse which angle goes with which triangle. Mark the angles clearly and use tan ratios step by step.

🎯 Exam Tip: Always write the final answer with correct units (m/sec). Show the tan 60° = √3 and tan 30° = 1/√3 values clearly. Examiners check these values.

Answer 45.

[Diagram: A right triangle showing a cliff AB with height 450 m. Point C is the initial position of a boat, and point D is the position after 3 minutes. Angle ACB is 30° and angle ADB is 60°.]

Let AB be the cliff. Then, AB = 450 m
Initial position of boat is C, which changes to D after 3 minutes.

In ΔADB
\( \frac{AB}{DB} = \tan 60° \)
\( \frac{450}{DB} = \sqrt{3} \)
\( DB = \frac{450}{\sqrt{3}} \)

In ΔABC
\( \frac{AB}{BC} = \tan 30° \)
\( \frac{450}{BD + DC} = \frac{1}{\sqrt{3}} \)
\( 450\sqrt{3} = BD + DC \)
\( 450\sqrt{3} = \frac{450}{\sqrt{3}} + DC \)
\( DC = 450\sqrt{3} - \frac{450}{\sqrt{3}} = 450\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) \)
\( = \frac{900}{\sqrt{3}} = \frac{900}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = 300\sqrt{3} \)

Time taken by car to travel DC distance \( \left(i.e., 300\sqrt{3}\right) = 3 \) minutes
Time taken by car to travel DB distance \( \left(i.e., \frac{450}{\sqrt{3}}\right) = \frac{3}{300\sqrt{3}} \times \frac{450}{\sqrt{3}} = \frac{450}{300} = 1.5 \)

Thus, the time it will take to reach the shore is 1 min 30 secs.

Speed of the boat = \( \frac{\text{Distance}}{\text{Time}} \)
\( = \frac{300\sqrt{3}}{3} = 100\sqrt{3} = 100 \times 1.732 = 173.2 \) m/min
\( = \frac{173.2}{60} \) m/sec = 2.9 m/sec
In simple words: The boat travels from C to D in 3 minutes. We found this distance using cliff height and angles. From D to shore (B) takes 1.5 more minutes.

📝 Teacher's Note: Students often forget to convert minutes to seconds. Practice both time calculations and unit conversions. Show that tan 30° and tan 60° are reciprocals.

🎯 Exam Tip: Write time answers in both minutes and seconds format. Always convert speed to m/sec at the end. Show all calculation steps clearly for full marks.

Answer 46.

[Diagram: A right triangle showing a tower AB. Point C is the initial position of a ship, and point D is the position after 3 minutes. Angle ACB is 30° and angle ADB is 60°.]

Let AB be the tower.
Initial position of ship is C, which changes to D after 3 minutes.

In ΔADB
\( \frac{AB}{DB} = \tan 60° \)
\( \frac{AB}{DB} = \sqrt{3} \)
\( DB = \frac{AB}{\sqrt{3}} \)

In ΔABC
\( \frac{AB}{BC} = \tan 30° \)
\( \frac{AB}{BD + DC} = \frac{1}{\sqrt{3}} \)
\( AB\sqrt{3} = BD + DC \)
\( AB\sqrt{3} = \frac{AB}{\sqrt{3}} + DC \)
\( DC = AB\sqrt{3} - \frac{AB}{\sqrt{3}} = AB\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) \)
\( = \frac{2AB}{\sqrt{3}} \)

Time taken by car to travel DC distance \( \left(i.e., \frac{2AB}{\sqrt{3}}\right) = 3 \) minutes
Time taken by car to travel DB distance \( \left(i.e., \frac{AB}{\sqrt{3}}\right) = \frac{3}{\frac{2AB}{\sqrt{3}}} \times \frac{AB}{\sqrt{3}} = \frac{3}{2} = 1 \) min 30 secs

Thus, the total time taken is 3 minutes + 1 minute 30 seconds = 4 minutes 30 seconds.
In simple words: The ship takes 3 minutes to go from C to D. Then it takes 1.5 more minutes to reach the tower from D. Total time is 4 minutes 30 seconds.

📝 Teacher's Note: This problem is about finding total travel time. First find the ratio of distances CD to DB, then use the given time to find the remaining time.

🎯 Exam Tip: Add the two time periods carefully. Write the final answer clearly as "4 minutes 30 seconds" or "4.5 minutes". Show the ratio calculation step.

Answer 47.

[Diagram: A right triangle showing a tower AB. Point A is the position of a man on top. Points C and D are two positions of a truck on the ground. Angles α and β are marked.]

In the figure, AB is the tower. A is the position of the man. C and D are the two positions of the truck.

Let the speed of the truck be x m/sec
Distance CD = speed × time = 600x

In right triangle ABC,
\( \tan α = \frac{h}{BC} \)
It is given that \( \tan α = \frac{1}{\sqrt{5}} \)
\( BC = h\sqrt{5} \) ... (1)

In right triangle ABD,
\( \tan β = \frac{h}{BD} \)
It is given that \( \tan β = \sqrt{5} \)
\( h = \sqrt{5}BD \)

Now, CD = BC - BD
600x = 5BD - BD
BD = 150x

Time taken = \( \frac{150x}{x} = 150 \) seconds

Thus, the time taken by the truck to reach the tower is 150 sec = 2 min 30 sec.
In simple words: We used the two angle values to find the distances. Then we used the truck's movement to calculate how long it takes to reach the tower.

📝 Teacher's Note: Students struggle with this type because there are two triangles. Draw both triangles separately first. Then show how CD = BC - BD.

🎯 Exam Tip: Write the tan values clearly at the start. Show that CD is the difference of the two base lengths. Convert seconds to minutes and seconds in the final answer.

Answer 48.

[Diagram: A quadrilateral showing an aeroplane's path. Point A is the observation point on ground. Points B and C are two positions of the aeroplane at height 3000 m. Angles 45° and 30° are marked.]

Let A be the point of observation on the ground and B and C be the two positions of aeroplane. Let BL = CM = 3000 m.

In ΔALB,
\( \tan 45° = \frac{BL}{AL} \)
\( 1 = \frac{3000}{AL} \)
\( AL = 3000 \)

In ΔAMC,
\( \tan 30° = \frac{MC}{AM} \)
\( \frac{1}{\sqrt{3}} = \frac{3000}{3000 + LM} \)
\( 3000\sqrt{3} = (3000 + LM) \)
\( LM = 3000(\sqrt{3} - 1) \)
\( BC = 3000(\sqrt{3} - 1) \)

Now, time taken to travel distance BC = 15 seconds
Speed of the aeroplane = \( \frac{\text{Distance}}{\text{Time}} = \frac{3000(\sqrt{3} - 1)}{15} = 200 \times 0.732 = 146.4 \)

Thus, the speed of the aeroplane is 146.4 m/sec
\( = 146.4 \times \frac{1000}{3600} \) km/hr = \( 146.4 \times 3.6 \) km/hr = 527.04 km/hr
In simple words: The plane moves from B to C in 15 seconds. We used the two angles to find this distance BC. Then we divided distance by time to get speed.

📝 Teacher's Note: Students often forget that tan 45° = 1. Also remind them that √3 ≈ 1.732, so (√3 - 1) ≈ 0.732. This makes calculations easier.

🎯 Exam Tip: Show both speed units - m/sec and km/hr. Use the conversion factor 3.6 to go from m/sec to km/hr. Write √3 = 1.732 clearly in your working.

Answer 49.
Answer:

[Diagram: A tower PQ with height h, showing angles of elevation from two points X and Y. Point X is at ground level with angle 60°, point Y is 40m above X with angle 45°. The tower and observation points form right triangles.]

In the figure, PQ is the tower.

Step 1: Find relationship in triangle PQX.
In ΔPQX,
\( \frac{h}{x} = \tan 60° = \sqrt{3} \)
\( \implies h = \sqrt{3}x \) ...(1)

Step 2: Find relationship in triangle QRY.
In ΔQRY,
\( \frac{h - 40}{x} = \tan 45° = 1 \)
\( \implies h = 40 + x \) ...(2)

Step 3: Solve equations (1) and (2).
From (1) and (2),
\( \sqrt{3}x = 40 + x \)
\( \implies (\sqrt{3} - 1)x = 40 \)
\( \implies x = \frac{40}{\sqrt{3} - 1} = \frac{40(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{40(\sqrt{3} + 1)}{2} = 20(\sqrt{3} + 1) \)

Step 4: Calculate height of tower.
\( \therefore h = 40 + 20(\sqrt{3} + 1) = 40 + 20\sqrt{3} + 20 = 60 + 20\sqrt{3} = 20(3 + \sqrt{3}) = 20 \times 4.732 = 94.64 \)

Step 5: Find distance XQ.
Again, in ΔPQX,
\( \frac{h}{XQ} = \sin 60° = \frac{\sqrt{3}}{2} \)
\( \implies XQ = \frac{2h}{\sqrt{3}} = \frac{2 \times 94.64}{\sqrt{3}} = 109.3 \text{ m} \)

Thus, the height of the tower PQ is 94.64 m.

In simple words: We used two different viewing angles to find the tower height. From the lower point, we see the tower at 60°. From 40m higher, we see it at 45°. Using these angles, we found the tower is about 95 meters tall.

📝 Teacher's Note: Draw a clear diagram first. Show students how we make two right triangles with the same tower height. The key is that both triangles share the same tower height h.

🎯 Exam Tip: Always label your diagram clearly. Write "Let height = h" at the start. Show each step of calculation clearly. Don't forget to rationalize the denominator when needed.

Answer 50.
Answer:

[Diagram: A cloud C above a lake surface AB. Point P is the observation point at height h above the lake. The cloud appears at position C and its reflection appears at C'. Angles of elevation and depression are marked as α and β respectively.]

Let AB be the surface of the lake and let P be an point of observation such that AP = h meters. Let C be the position of the cloud and C' be its reflection in the lake. Then ∠CPM = α and ∠MPC' = β. Let CM = x. Then, CB = CM + MB = CM + PA = x + h

Step 1: Find relationship in triangle CPM.
In Δ CPM,
\( \tan α = \frac{CM}{PM} \)
\( \implies \tan α = \frac{x}{AB} \) [∵ PM = AB]
\( \implies AB = x \cot α \) ... (1)

Step 2: Find relationship in triangle PMC'.
In Δ PMC',
\( \tan β = \frac{C'M}{PM} \)
\( \implies \tan β = \frac{x + 2h}{AB} \)
\( \implies AB = (x + 2h) \cot β \) ... (2)

Step 3: Solve equations (1) and (2).
From (1) and (2),
\( x \cot α = (x + 2h) \cot β \)
\( \implies x\left(\frac{1}{\tan α} - \frac{1}{\tan β}\right) = \frac{2h}{\tan β} \)
\( \implies x\left(\frac{\tan β - \tan α}{\tan α \tan β}\right) = \frac{2h}{\tan β} \)
\( \implies x = \frac{2h \tan α}{\tan β - \tan α} \)

Step 4: Find height of cloud.
Again, in Δ CPM,

In simple words: We use the reflection of the cloud in the lake water. The cloud and its reflection help us make two triangles. We can then find how high the cloud is above the lake.

📝 Teacher's Note: Use a mirror to show students how reflections work. The reflection appears as far below the surface as the object is above it. This makes the total distance = original height + 2 times the gap.

🎯 Exam Tip: Remember that the reflected image is at the same distance below the surface as the object is above it. Always write the formula for reflection distance clearly.

Answer 51.
Answer:

[Diagram: An aeroplane at point P flying horizontally. Two observation points A and Q on the ground, with Q being x miles from A. Angles of elevation from both points are marked as α and β respectively.]

Let P Q be h
QB be x
Given : AB = 1 mile
QB = x
AQ = (1- x) mile
in ΔPAQ

Step 1: Find relationship in triangle PAQ.
\( \tan α = \frac{PQ}{AQ} \)
\( \tan α = \frac{h}{1-x} \)
\( 1 - x = \frac{h}{\tan α} \) ...........(1)

Step 2: Find relationship in triangle PQB.
In ΔPQB
\( \tan β = \frac{h}{x} \)
\( x = \frac{h}{\tan β} \)

Step 3: Substitute and solve.
Substitute for x in equation (1)
\( 1 = \frac{h}{\tan β} + \frac{h}{\tan α} \)
\( 1 = h\left(\frac{1}{\tan β} + \frac{1}{\tan α}\right) \)
\( \frac{1}{h} = \frac{\tan β + \tan α}{\tan β \tan α} \)

Thus, the height in miles of aeroplane above the road is \( \frac{\tan α \tan β}{\tan α + \tan β} \).

In simple words: We have an airplane flying straight above a road. Two people on the road see it at different angles. Using these two angles, we can find how high the plane is flying.

📝 Teacher's Note: Draw the horizontal line for the airplane path clearly. Show students that the two observation points are 1 mile apart. The key is to set up two right triangles with the same height.

🎯 Exam Tip: Always write the final answer in the form asked. Here the answer is a fraction with tan functions. Don't try to simplify it to decimal unless asked.

Answer 52.
Answer:

[Diagram: A ladder that can slide between two positions - CP and DW. The ladder moves from wall DC (height a) to wall PW (height b). Angles α and β are marked for the two positions.]

Let CP and DW be the two positions of the ladder such that CP = DW = l (say).

CD = a, PW = b, ∠ACP = α and ∠ADW = β

Step 1: Find relationship in triangle APC.
In ΔAPC,
\( \frac{AC}{CP} = \cos α \implies AC = x \cos α \) ...(I)

Step 2: Find relationship in triangle ADW.
In ΔADW,
\( \frac{AD}{DW} = \cos β \implies \frac{AC + CD}{DW} = \cos β \)
\( \implies \frac{x \cos α + a}{x} = \cos β \) [using (I)]
\( \implies x = \frac{a}{\cos β - \cos α} \) ...(II)

Step 3: Find AP using triangle APC.
Again In ΔAPC, \( \frac{AP}{CP} = \sin α \)
\( \implies AP = x \sin α = \frac{a \sin α}{(\cos β - \cos α)} \) ...(III) [Using (II)]

Step 4: Find AW using triangle ADW.
Again In ΔADW, \( \frac{AW}{DW} = \sin β \)
\( \implies AW = x \sin β = \frac{a \sin β}{(\cos β - \cos α)} \) ...(IV)

In simple words: A ladder can slide between two walls of different heights. We use the angles it makes in both positions to find the distance between the walls and other measurements.

📝 Teacher's Note: Use two books of different heights and a ruler to demonstrate this problem. Show how the same ladder (ruler) makes different angles when moved between the two walls (books).

🎯 Exam Tip: Draw both positions of the ladder clearly. Label all the given information first. Remember that the ladder length remains the same in both positions.

Answer 53.
Answer:

[Diagram: A lighthouse AB of height h meters. Two ships at positions C and D on the sea. AC = x and AD = y. Angles of depression α and β are marked from the lighthouse to the ships.]

Let AB be the lighthouse of height h m. Let AC = x and AD = y.

Step 1: Find relationship in triangle CAB.
In ΔCAB,
\( \frac{AB}{AC} = \tan α \)
\( \tan α = \frac{h}{x} \)
\( x = \frac{h}{\tan α} \) ... (i)

Step 2: Find relationship in triangle DAB.
In ΔDAB,
\( \frac{AB}{AD} = \tan β \)
\( \tan α = \frac{h}{y} \)
\( y = \frac{h}{\tan β} \) ... (ii)

Step 3: Find distance between ships.
Distance between the ships = x + y
\( = \frac{h}{\tan α} + \frac{h}{\tan β} \)
\( = h\left(\frac{\tan β + \tan α}{\tan α \tan β}\right) \)

Therefore, the distance between the ships is \( h\left(\frac{\tan β + \tan α}{\tan α \tan β}\right) \) meters.

In simple words: A lighthouse keeper looks down at two ships in the sea. Using the angles at which he sees them, we can find how far apart the ships are from each other.

📝 Teacher's Note: Explain that angles of depression are measured downward from the horizontal. Use a flashlight to show students how someone at height looks down at objects below.

🎯 Exam Tip: Remember that angles of depression from the lighthouse equal angles of elevation from the ships. Always draw the horizontal line from the lighthouse first, then mark the depression angles below it.