CBSE Class 10 Mathematics Some Applications of Trigonometry MCQs Set 10

Answer: (c) 864

Explanation:
1. Horizontal distances: \( \frac{3600\sqrt{3}}{\sqrt{3}} \) = 3600 m and \( 3600\sqrt{3} \times \sqrt{3} \) = 10800 m.
2. Distance flown = 7200 m in 30 s, so speed = 240 m/s.
3. 240 × 18/5 = 864 km/h.

Teacher's Note:
1. Multiply m/s by 18/5 to get km/h.
2. The two horizontal distances are h cot 60° and h cot 30°.

Answer: (b) 1039.2 m

Explanation:
1. 432 km/h = 120 m/s, so the distance flown is 1200 m.
2. \( \frac{2h}{\sqrt{3}} = 1200 \), so h = 600\( \sqrt{3} \) = 1039.2 m.

Teacher's Note:
1. Distance flown = h(cot 30° − cot 60°).
2. The 648 km/h version gives 1558.8 m.

Answer: (a) 2598 m

Explanation:
1. 720 km/h = 200 m/s, so distance = 200 × 15 = 3000 m.
2. \( \frac{2h}{\sqrt{3}} = 3000 \), so h = 1500\( \sqrt{3} \) = 2598 m.

Teacher's Note:
1. Convert the speed to m/s first.
2. 1500 × 1.732 = 2598.

Answer: (d) 2 m

Explanation:
1. Let the pedestal be h and the distance be x. tan 45° gives x = h, and tan 60° gives h + 1.46 = x\( \sqrt{3} \) = 1.73h.
2. 0.73h = 1.46, so h = 2 m.

Teacher's Note:
1. Both angles use the same horizontal distance.
2. tan 45° = 1.

Answer: (c) 13.5 m

Explanation:
1. Let the tower be H and the distance to the building d. From the bottom: d = \( \frac{H}{\sqrt{3}} \).
2. From the top of the building: H − 9 = d tan 30° = \( \frac{H}{3} \).
3. \( \frac{2H}{3} \) = 9, so H = 13.5 m (and d ≈ 7.8 m).

Teacher's Note:
1. The top of the building is H − 9 below the tower's top level.
2. Check: 4.5 m below the top of the tower is 13.5 × 1/3.

Answer: (b) 200 m

Explanation:
1. Let the helicopter be at height H. It is H − 100 above the observer and its reflection is H + 100 below the observer's level.
2. \( \frac{H - 100}{x} = \frac{1}{\sqrt{3}} \) and \( \frac{H + 100}{x} = \sqrt{3} \), so H + 100 = 3(H − 100).
3. 2H = 400, so H = 200 m.

Teacher's Note:
1. The reflection is as far below the lake surface as the object is above it.
2. Divide the two equations to remove x.

Answer: (a) 88.3 m

Explanation:
1. The balloon moves \( 29\sqrt{3} \times 2 = 58\sqrt{3} \) m horizontally.
2. With h as its height above the eye: h cot 30° − h cot 60° = \( \frac{2h}{\sqrt{3}} \) = \( 58\sqrt{3} \), so h = 87 m.
3. Height above the ground = 87 + 1.3 = 88.3 m.

Teacher's Note:
1. Add the boy's eye level at the end.
2. Horizontal distance = h cot θ.

Answer: (d) 18.93 m

Explanation:
1. Let the tall building be H, the distance d = H (tan 45° = 1).
2. H − 8 = d tan 30° = \( \frac{H}{\sqrt{3}} \), so H(\( \sqrt{3} \) − 1) = 8\( \sqrt{3} \).
3. H = \( \frac{8\sqrt{3}}{\sqrt{3} - 1} \) = 4(3 + \( \sqrt{3} \)) = 18.93 m (the buildings are also 18.93 m apart).

Teacher's Note:
1. Rationalise using \( \sqrt{3} + 1 \).
2. The distance between the buildings equals H.

Answer: (c) 34.64 m

Explanation:
1. From the bottom of the post: tan 60° = \( \frac{60}{d} \), so d = \( \frac{60}{\sqrt{3}} \) = 20\( \sqrt{3} \) = 34.64 m.
2. (The post is 60 − d tan 30° = 60 − 20 = 40 m high.)

Teacher's Note:
1. The foot of the post is at depression 60°.
2. Rationalise 60/√3 = 20√3.

Answer: (b) 20 m

Explanation:
1. The distance between them is d = \( \frac{60}{\tan 60°} = 20\sqrt{3} \) m.
2. The tower rises above the building top by d tan 30° = 20 m.

Teacher's Note:
1. The part of the tower above the building is found from the elevation angle.
2. The tower is 80 m high.

Answer: (a) 45°

Explanation:
1. The height of the tower above the eye is 22 − 1.5 = 20.5 m.
2. tan θ = \( \frac{20.5}{20.5} \) = 1, so θ = 45°.

Teacher's Note:
1. Subtract the observer's height first.
2. tan 45° = 1.

Answer: (d) 400 m

Explanation:
1. Let the cloud be at height H above the lake. It is H − 200 above the observer; its reflection is H + 200 below the observer.
2. H + 200 = 3(H − 200), so 2H = 800 and H = 400 m.

Teacher's Note:
1. Use tan 60° = 3 × tan 30° to relate the two triangles.
2. The reflection is at depth H below the lake surface.

Answer: (c) 0.75(\( \sqrt{3} \) + 1) m

Explanation:
1. Let the pedestal be h and the distance x. tan 30° gives x = h\( \sqrt{3} \); tan 45° gives h + 1.5 = x = h\( \sqrt{3} \).
2. h(\( \sqrt{3} \) − 1) = 1.5, so h = \( \frac{1.5}{\sqrt{3} - 1} \) = 0.75(\( \sqrt{3} \) + 1) ≈ 2.05 m.

Teacher's Note:
1. Rationalise by multiplying with \( \sqrt{3} + 1 \).
2. Both angles use the same horizontal distance.

Answer: (b) 129.9 m

Explanation:
1. Let the height be h. h cot 30° − h cot 60° = 150, so \( \frac{2h}{\sqrt{3}} \) = 150.
2. h = 75\( \sqrt{3} \) = 129.9 m.

Teacher's Note:
1. The distance moved is the difference of the two horizontal distances.
2. 75 × 1.732 = 129.9.

Answer: (a) 36.6 m

Explanation:
1. Let the height be h. The ships are h cot 30° and h cot 45° from the foot, on opposite sides.
2. h(\( \sqrt{3} \) + 1) = 100, so h = \( \frac{100}{2.732} \) = 36.6 m.

Teacher's Note:
1. Opposite sides: add the two distances.
2. 100 ÷ 2.732 = 36.6.