CBSE Class 10 Mathematics Some Applications of Trigonometry MCQs Set 09

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 09 Some Applications of Trigonometry

Review structured MCQ sets for Class 10 Mathematics Chapter 09 Some Applications of Trigonometry. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Chapter 09 Some Applications of Trigonometry MCQs for Class 10 Mathematics

View or download the dedicated Chapter 09 Some Applications of Trigonometry MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: A circus artist climbs a 15 m long rope that is tied from the top of a vertical pole to the ground. If the rope makes an angle of 45° with the ground, the height of the pole is:
ABC15 m (a) 15 m
(b) \( 15\sqrt{2} \) m
(c) \( \frac{15}{\sqrt{3}} \) m
(d) \( \frac{15}{\sqrt{2}} \) m
Show Answer & Explanation

Answer: (d) \( \frac{15}{\sqrt{2}} \) m

Explanation:
1. sin 45° = h/15, so h = 15/√2 m.

Teacher's Note:
1. The rope is the hypotenuse; the pole is opposite the angle.
2. sin 45° = 1/√2.

Question 2: A circus artist climbs a 15 m long rope that is tied from the top of a vertical pole to the ground. If the angle stays 45°, the distance between the artist's starting point (foot of rope) and the pole at ground level is:
ABC15 m (a) \( \frac{15}{\sqrt{2}} \) m
(b) \( 15\sqrt{2} \) m
(c) 15 m
(d) \( \frac{15}{\sqrt{3}} \) m
Show Answer & Explanation

Answer: (a) \( \frac{15}{\sqrt{2}} \) m

Explanation:
1. cos 45° = x/15, so x = 15/√2 m.

Teacher's Note:
1. At 45° the base equals the height.
2. cos 45° = 1/√2.

Question 3: A circus artist climbs a 15 m long rope that is tied from the top of a vertical pole to the ground. If the angle made by the rope with the ground changes to 30°, the height of the pole is:
(a) 2.5 m
(b) 5 m
(c) 7.5 m
(d) 10 m
Show Answer & Explanation

Answer: (c) 7.5 m

Explanation:
1. sin 30° = h/15, so h = 15/2 = 7.5 m.

Teacher's Note:
1. sin 30° = 1/2.
2. Half of the rope length is the height.

Question 4: A circus artist climbs a 15 m long rope that is tied from the top of a vertical pole to the ground. If the angle is 30° and 3 m of the rope is cut off (so the rope is 12 m), the height of the pole is:
(a) 2 m
(b) 4 m
(c) 5 m
(d) 6 m
Show Answer & Explanation

Answer: (d) 6 m

Explanation:
1. With a 12 m rope, h = 12 sin 30° = 6 m.

Teacher's Note:
1. The new rope length is 15 − 3 = 12 m.
2. Apply sin 30° = 1/2 again.

Question 5: An electrician has to repair a fault at a point B on a pole AD of height 8 m, where B is 2 m below the top A. The foot of the ladder is C and it makes an angle of 60° with the ground. The length BD is:
ABCD60°2 m (a) 10 m
(b) 6 m
(c) 4 m
(d) 2 m
Show Answer & Explanation

Answer: (b) 6 m

Explanation:
1. BD = AD − AB = 8 − 2 = 6 m.

Teacher's Note:
1. BD is the part of the pole below the repair point.
2. Subtract 2 from 8.

Question 6: An electrician has to repair a fault at a point B on a pole AD of height 8 m, where B is 2 m below the top A. The foot of the ladder is C and it makes an angle of 60° with the ground. The length of the ladder BC is:
ABCD60°2 m (a) \( 4\sqrt{3} \) m
(b) \( 6\sqrt{3} \) m
(c) \( 8\sqrt{3} \) m
(d) \( 2\sqrt{3} \) m
Show Answer & Explanation

Answer: (a) \( 4\sqrt{3} \) m

Explanation:
1. sin 60° = BD/BC, so BC = 6/(√3/2) = 12/√3 = 4√3 m.

Teacher's Note:
1. The ladder is the hypotenuse.
2. Rationalise 12/√3.

Question 7: An electrician has to repair a fault at a point B on a pole AD of height 8 m, where B is 2 m below the top A. The foot of the ladder is C and it makes an angle of 60° with the ground. The distance DC between the foot of the ladder and the pole is:
ABCD60°2 m (a) \( 4\sqrt{3} \) m
(b) \( 6\sqrt{3} \) m
(c) \( 8\sqrt{3} \) m
(d) \( 2\sqrt{3} \) m
Show Answer & Explanation

Answer: (d) \( 2\sqrt{3} \) m

Explanation:
1. tan 60° = BD/DC, so DC = 6/√3 = 2√3 m.

Teacher's Note:
1. Use tan for the two legs.
2. 6/√3 = 2√3.

Question 8: An electrician has to repair a fault at a point B on a pole AD of height 8 m, where B is 2 m below the top A. The foot of the ladder is C and it makes an angle of 60° with the ground. The measure of ∠BCD when BD and CD are equal is:
(a) 30°
(b) 45°
(c) 60°
(d) 75°
Show Answer & Explanation

Answer: (b) 45°

Explanation:
1. tan ∠BCD = BD/CD = 1, so ∠BCD = 45°.

Teacher's Note:
1. Equal legs make an isosceles right triangle.
2. tan 45° = 1.

Question 9: An electrician has to repair a fault at a point B on a pole AD of height 8 m, where B is 2 m below the top A. The foot of the ladder is C and it makes an angle of 60° with the ground. For the 60° ladder, the measure of ∠DBC is:
(a) 15°
(b) 60°
(c) 30°
(d) 45°
Show Answer & Explanation

Answer: (c) 30°

Explanation:
1. In triangle BDC, ∠D = 90° and ∠C = 60°, so ∠B = 180° − 90° − 60° = 30°.

Teacher's Note:
1. Angle sum of a triangle is 180°.
2. The two acute angles are complementary.

Question 10: Shravan stands at the top of a building and sees a car approaching its foot at uniform speed at an angle of depression of 30°. Six seconds later the angle of depression is 60° and the car is then 25 m from the building's foot. The height of the building is:
60°30°ABCD25 m (a) \( \frac{25}{\sqrt{3}} \) m
(b) 50 m
(c) \( 25\sqrt{3} \) m
(d) 25 m
Show Answer & Explanation

Answer: (c) \( 25\sqrt{3} \) m

Explanation:
1. tan 60° = h/25, so h = 25√3 m.

Teacher's Note:
1. Use the second position, where the distance is known.
2. tan 60° = √3.

Question 11: Shravan stands at the top of a building and sees a car approaching its foot at uniform speed at an angle of depression of 30°. Six seconds later the angle of depression is 60° and the car is then 25 m from the building's foot. The distance between the two positions of the car is:
60°30°ABCD25 m (a) 40 m
(b) 50 m
(c) 60 m
(d) 75 m
Show Answer & Explanation

Answer: (b) 50 m

Explanation:
1. At the first position, tan 30° = 25√3/BD, so BD = 25√3 × √3 = 75 m.
2. Distance covered = 75 − 25 = 50 m.

Teacher's Note:
1. Find the first distance from the 30° angle.
2. Subtract the two distances.

Question 12: Shravan stands at the top of a building and sees a car approaching its foot at uniform speed at an angle of depression of 30°. Six seconds later the angle of depression is 60° and the car is then 25 m from the building's foot. The total time taken by the car to reach the foot of the building from the starting point is:
(a) 4 s
(b) 3 s
(c) 6 s
(d) 9 s
Show Answer & Explanation

Answer: (d) 9 s

Explanation:
1. 50 m takes 6 s, so 25 m takes 3 s.
2. Total time = 6 + 3 = 9 s.

Teacher's Note:
1. The speed is uniform, so time is proportional to distance.
2. Total distance is 75 m.

Question 13: Shravan stands at the top of a building and sees a car approaching its foot at uniform speed at an angle of depression of 30°. Six seconds later the angle of depression is 60° and the car is then 25 m from the building's foot. The distance of the observer from the car when the angle of depression is 60° is:
(a) 25 m
(b) 45 m
(c) 50 m
(d) 75 m
Show Answer & Explanation

Answer: (c) 50 m

Explanation:
1. cos 60° = 25/AC, so AC = 25/(1/2) = 50 m.

Teacher's Note:
1. The line of sight is the hypotenuse.
2. cos 60° = 1/2.

Question 14: Shravan stands at the top of a building and sees a car approaching its foot at uniform speed at an angle of depression of 30°. Six seconds later the angle of depression is 60° and the car is then 25 m from the building's foot. The angle of elevation increases when:
(a) the point of observation moves towards the object
(b) the point of observation moves away from the object
(c) the object moves away from the observer
(d) none of the above
Show Answer & Explanation

Answer: (a) the point of observation moves towards the object

Explanation:
1. Moving closer to the object makes the horizontal distance smaller, so tan(elevation) = height/distance becomes larger.
2. The angle of elevation increases when the observation point moves towards the object.

Teacher's Note:
1. Height stays constant, so a smaller distance means a larger angle.
2. Use tan θ = height/distance.

Question 15: A boy stands on top of a lighthouse of height 100 m. Boats A and B approach the lighthouse from opposite directions. The angle of depression of boat A is 45° and of boat B is 30°. P is the top of the lighthouse and D is its foot. The angle ∠PAD (angle of the line of sight with the ground at boat A) is:
PABD45°30° (a) 30°
(b) 45°
(c) 60°
(d) 90°
Show Answer & Explanation

Answer: (b) 45°

Explanation:
1. The angle of depression of A is 45°. By alternate interior angles, ∠PAD = 45°.

Teacher's Note:
1. The angle at the boat equals the angle of depression.
2. Parallel lines give equal alternate angles.

Question 16: A boy stands on top of a lighthouse of height 100 m. Boats A and B approach the lighthouse from opposite directions. The angle of depression of boat A is 45° and of boat B is 30°. P is the top of the lighthouse and D is its foot. If the angle of depression of B is 30°, then ∠PBD is also 30°. Why?
PABD45°30° (a) vertically opposite angles
(b) alternate interior angles
(c) alternate exterior angles
(d) corresponding angles
Show Answer & Explanation

Answer: (b) alternate interior angles

Explanation:
1. The horizontal line through P and the ground are parallel, and PB is a transversal, so the angles are alternate interior angles.

Teacher's Note:
1. Name the two parallel lines first.
2. Alternate interior angles lie between the parallels on opposite sides of the transversal.

Question 17: A boy stands on top of a lighthouse of height 100 m. Boats A and B approach the lighthouse from opposite directions. The angle of depression of boat A is 45° and of boat B is 30°. P is the top of the lighthouse and D is its foot. The distance AD of boat A from the foot of the lighthouse is:
PABD45°30° (a) 90 m
(b) 60 m
(c) 100 m
(d) 80 m
Show Answer & Explanation

Answer: (c) 100 m

Explanation:
1. tan 45° = 100/AD, so AD = 100 m.

Teacher's Note:
1. At 45° the base equals the height.
2. tan 45° = 1.

Question 18: A boy stands on top of a lighthouse of height 100 m. Boats A and B approach the lighthouse from opposite directions. The angle of depression of boat A is 45° and of boat B is 30°. P is the top of the lighthouse and D is its foot. The distance BD of boat B from the foot of the lighthouse is:
PABD45°30° (a) 50 m
(b) 100 m
(c) \( \frac{100}{\sqrt{3}} \) m
(d) \( 100\sqrt{3} \) m
Show Answer & Explanation

Answer: (d) \( 100\sqrt{3} \) m

Explanation:
1. tan 30° = 100/BD, so BD = 100√3 m.

Teacher's Note:
1. BD = height / tan 30°.
2. 1/tan 30° = √3.

Question 19: A boy stands on top of a lighthouse of height 100 m. Boats A and B approach the lighthouse from opposite directions. The angle of depression of boat A is 45° and of boat B is 30°. P is the top of the lighthouse and D is its foot. The distance of boat A from the top of the lighthouse (line of sight PA) is:
PABD45°30° (a) \( 100\sqrt{2} \) m
(b) \( \frac{100}{\sqrt{2}} \) m
(c) 50 m
(d) 100 m
Show Answer & Explanation

Answer: (a) \( 100\sqrt{2} \) m

Explanation:
1. sin 45° = 100/PA, so PA = 100√2 m.

Teacher's Note:
1. PA is the hypotenuse of the right triangle PDA.
2. sin 45° = 1/√2.

Question 20: A statue stands on a pedestal. From a point A on the ground, the angle of elevation of the top of the pedestal is 45° and of the top of the statue is 60°. C is the foot of the pedestal, D its top and B the top of the statue. If the height of the pedestal is 20 m, the distance AC between the point A and the foot of the pedestal is:
45°60°ACDBpedestal CDstatue DB (a) 20 m
(b) 40 m
(c) 60 m
(d) 80 m
Show Answer & Explanation

Answer: (a) 20 m

Explanation:
1. tan 45° = CD/AC = 1, so AC = CD = 20 m.

Teacher's Note:
1. At 45° the horizontal distance equals the height.
2. The pedestal's top is at 45°.

Question 21: A statue stands on a pedestal. From a point A on the ground, the angle of elevation of the top of the pedestal is 45° and of the top of the statue is 60°. C is the foot of the pedestal, D its top and B the top of the statue. If the pedestal is 20 m high, the height of the statue is:
45°60°ACDBpedestal CDstatue DB (a) \( 20\sqrt{3} \) m
(b) \( 20(\sqrt{3}-1) \) m
(c) \( 20(\sqrt{3}+1) \) m
(d) \( \frac{20}{\sqrt{3}} \) m
Show Answer & Explanation

Answer: (b) \( 20(\sqrt{3}-1) \) m

Explanation:
1. tan 60° = BC/AC, so BC = 20√3.
2. Statue height DB = BC − CD = 20√3 − 20 = 20(√3 − 1) m.

Teacher's Note:
1. The 60° line of sight reaches the top of the statue, i.e. the total height.
2. Subtract the pedestal's height.

Question 22: A statue stands on a pedestal. From a point A on the ground, the angle of elevation of the top of the pedestal is 45° and of the top of the statue is 60°. C is the foot of the pedestal, D its top and B the top of the statue. If the height of the statue is 1.6 m, the height of the pedestal is:
45°60°ACDBpedestal CDstatue DB (a) \( 1.6\sqrt{3} \) m
(b) \( 0.8(\sqrt{3}-1) \) m
(c) \( 1.6(\sqrt{3}+1) \) m
(d) \( 0.8(\sqrt{3}+1) \) m
Show Answer & Explanation

Answer: (d) \( 0.8(\sqrt{3}+1) \) m

Explanation:
1. Let the pedestal be x m. Then AC = x and tan 60° = (x + 1.6)/x.
2. √3x = x + 1.6, so x = 1.6/(√3 − 1) = 0.8(√3 + 1) m.

Teacher's Note:
1. Set up the equation for the total height at 60°.
2. Rationalise 1.6/(√3 − 1).

Question 23: A statue stands on a pedestal. From a point A on the ground, the angle of elevation of the top of the pedestal is 45° and of the top of the statue is 60°. C is the foot of the pedestal, D its top and B the top of the statue. If the total height of the statue and pedestal is 39 m, the length of AC is:
45°60°ACDBpedestal CDstatue DB (a) 13 m
(b) \( 39\sqrt{3} \) m
(c) \( 13\sqrt{3} \) m
(d) \( \frac{39}{2} \) m
Show Answer & Explanation

Answer: (c) \( 13\sqrt{3} \) m

Explanation:
1. tan 60° = 39/AC, so AC = 39/√3 = 13√3 m.

Teacher's Note:
1. Total height is BC.
2. Rationalise 39/√3.

Question 24: A statue stands on a pedestal. From a point A on the ground, the angle of elevation of the top of the pedestal is 45° and of the top of the statue is 60°. C is the foot of the pedestal, D its top and B the top of the statue. If the pedestal is 35 m high, the length AD is:
45°60°ACDBpedestal CDstatue DB (a) \( 35\sqrt{2} \) m
(b) \( \frac{35}{\sqrt{2}} \) m
(c) 35 m
(d) 70 m
Show Answer & Explanation

Answer: (a) \( 35\sqrt{2} \) m

Explanation:
1. sin 45° = CD/AD, so AD = 35/(1/√2) = 35√2 m.

Teacher's Note:
1. AD is the hypotenuse of triangle ACD.
2. sin 45° = 1/√2.

Question 25: Two hoardings are placed on two poles PQ and RS of equal height 20 m, standing on either side of a road. From a point O on the road between them (not the mid-point), the angles of elevation of the tops Q and S are 60° and 30° respectively. (Take √3 = 1.73.) The length PO is:
PQRSO60°30°20 m20 m (a) 20 m
(b) \( 20\sqrt{3} \) m
(c) \( \frac{20}{\sqrt{3}} \) m
(d) none of these
Show Answer & Explanation

Answer: (c) \( \frac{20}{\sqrt{3}} \) m

Explanation:
1. tan 60° = 20/PO, so PO = 20/√3 m.

Teacher's Note:
1. The angle at O is 60°, the pole is the opposite side.
2. PO = 20/tan 60°.

Question 26: Two hoardings are placed on two poles PQ and RS of equal height 20 m, standing on either side of a road. From a point O on the road between them (not the mid-point), the angles of elevation of the tops Q and S are 60° and 30° respectively. (Take √3 = 1.73.) The length RO is:
PQRSO60°30°20 m20 m (a) 20 m
(b) \( 20\sqrt{3} \) m
(c) \( \frac{20}{\sqrt{3}} \) m
(d) none of these
Show Answer & Explanation

Answer: (b) \( 20\sqrt{3} \) m

Explanation:
1. tan 30° = 20/RO, so RO = 20√3 m.

Teacher's Note:
1. Same method with the 30° angle.
2. 1/tan 30° = √3.

Question 27: Two hoardings are placed on two poles PQ and RS of equal height 20 m, standing on either side of a road. From a point O on the road between them (not the mid-point), the angles of elevation of the tops Q and S are 60° and 30° respectively. (Take √3 = 1.73.) The width of the road is:
PQRSO60°30°20 m20 m (a) 31.23 m
(b) 35.68 m
(c) 39.73 m
(d) 46.24 m
Show Answer & Explanation

Answer: (d) 46.24 m

Explanation:
1. Width PR = PO + RO = 20/√3 + 20√3 = 80/√3.
2. With √3 = 1.73, PR = 80/1.73 = 46.24 m.

Teacher's Note:
1. Add the two horizontal distances.
2. Combine over a common denominator: 20/√3 + 60/√3.

Question 28: Two hoardings are placed on two poles PQ and RS of equal height 20 m, standing on either side of a road. From a point O on the road between them (not the mid-point), the angles of elevation of the tops Q and S are 60° and 30° respectively. (Take √3 = 1.73.) If the angle of elevation of the top of pole PQ is 45° instead, the length PO is:
(a) 20 m
(b) \( 20\sqrt{3} \) m
(c) \( \frac{20}{\sqrt{3}} \) m
(d) none of these
Show Answer & Explanation

Answer: (a) 20 m

Explanation:
1. tan 45° = 20/PO, so PO = 20 m.

Teacher's Note:
1. tan 45° = 1, so the base equals the height.
2. The pole's height does not change.

Question 29: A vertical tree AB is broken by a storm at a point C so that the top touches the ground at D, forming a right triangle BCD with the right angle at B. If the top of the broken tree touches the ground at a distance of 30 m from the foot of the tree, making an angle of inclination of 30°, the height of the remaining part of the tree is:
30°BCD30 m (a) 10 m
(b) \( 20\sqrt{3} \) m
(c) \( 10\sqrt{3} \) m
(d) 30 m
Show Answer & Explanation

Answer: (c) \( 10\sqrt{3} \) m

Explanation:
1. tan 30° = BC/BD = x/30, so x = 30/√3 = 10√3 m.

Teacher's Note:
1. BC is the part still standing.
2. Use tan for the two legs.

Question 30: A vertical tree AB is broken by a storm at a point C so that the top touches the ground at D, forming a right triangle BCD with the right angle at B. If the top of the broken tree touches the ground at a point whose distance from the foot equals the height of the remaining part, the angle of inclination of the broken part is:
(a) 30°
(b) 60°
(c) 45°
(d) none of these
Show Answer & Explanation

Answer: (c) 45°

Explanation:
1. BD = BC, so tan θ = BC/BD = 1 and θ = 45°.

Teacher's Note:
1. Equal legs give an isosceles right triangle.
2. tan 45° = 1.

Question 31: A vertical tree AB is broken by a storm at a point C so that the top touches the ground at D, forming a right triangle BCD with the right angle at B. A 6 m tall tree is broken by the wind so that its top touches the ground and makes an angle of 30° with the ground. At what height from the bottom of the tree is it broken?
(a) 2 m
(b) 4 m
(c) 8 m
(d) 10 m
Show Answer & Explanation

Answer: (a) 2 m

Explanation:
1. Let it break at height x. The broken part has length 6 − x and sin 30° = x/(6 − x).
2. 1/2 = x/(6 − x), so 6 − x = 2x and x = 2 m.

Teacher's Note:
1. The broken part is the hypotenuse of the right triangle.
2. sin 30° = opposite/hypotenuse.

Question 32: Aaradhya, sitting on a bridge across a river at a point C which is 6 m above the water level, observes the angles of depression of the banks A and B on opposite sides of the river to be 60° and 30° respectively. D is the point on the water level directly below C. (Take √3 = 1.73.) The distance AD is:
CABD60°30°6 m (a) 6 m
(b) \( 6\sqrt{3} \) m
(c) \( 2\sqrt{3} \) m
(d) \( \frac{6}{\sqrt{3}} \) m
Show Answer & Explanation

Answer: (c) \( 2\sqrt{3} \) m

Explanation:
1. tan 60° = CD/AD, so AD = 6/√3 = 2√3 m.

Teacher's Note:
1. Use the 60° angle for bank A.
2. Rationalise 6/√3.

Question 33: Aaradhya, sitting on a bridge across a river at a point C which is 6 m above the water level, observes the angles of depression of the banks A and B on opposite sides of the river to be 60° and 30° respectively. D is the point on the water level directly below C. (Take √3 = 1.73.) The distance BD is:
CABD60°30°6 m (a) 6 m
(b) \( 6\sqrt{3} \) m
(c) \( 2\sqrt{3} \) m
(d) \( \frac{6}{\sqrt{3}} \) m
Show Answer & Explanation

Answer: (b) \( 6\sqrt{3} \) m

Explanation:
1. tan 30° = CD/BD, so BD = 6√3 m.

Teacher's Note:
1. Use the 30° angle for bank B.
2. 1/tan 30° = √3.

Question 34: Aaradhya, sitting on a bridge across a river at a point C which is 6 m above the water level, observes the angles of depression of the banks A and B on opposite sides of the river to be 60° and 30° respectively. D is the point on the water level directly below C. (Take √3 = 1.73.) The width of the river is:
CABD60°30°6 m (a) 10.85 m
(b) 13.84 m
(c) 15.85 m
(d) 19.85 m
Show Answer & Explanation

Answer: (b) 13.84 m

Explanation:
1. Width AB = AD + BD = 2√3 + 6√3 = 8√3.
2. With √3 = 1.73, AB = 8 × 1.73 = 13.84 m.

Teacher's Note:
1. The two banks are on opposite sides, so add the distances.
2. 8√3 ≈ 13.84.

Question 35: Aaradhya, sitting on a bridge across a river at a point C which is 6 m above the water level, observes the angles of depression of the banks A and B on opposite sides of the river to be 60° and 30° respectively. D is the point on the water level directly below C. (Take √3 = 1.73.) If BD = 21 m (for the same 30° angle), the height of the bridge above the water is:
(a) 7 m
(b) 21 m
(c) \( 7\sqrt{3} \) m
(d) \( 21\sqrt{3} \) m
Show Answer & Explanation

Answer: (c) \( 7\sqrt{3} \) m

Explanation:
1. tan 30° = CD/21, so CD = 21/√3 = 7√3 m.

Teacher's Note:
1. Height = BD × tan 30°.
2. Rationalise 21/√3.

Question 36: Raju, 4 m tall, sees a pigeon on top of a 54 m pole. The angle of elevation of the pigeon from his eyes is 60°. The pigeon flies horizontally at constant height and after 8 seconds the angle of elevation from the same point is 45°. (Take √3 = 1.73.) The pigeon is therefore 54 − 4 = 50 m above his eye level. The distance of the first position of the pigeon from Raju's eyes is:
60°45°ACEBD (a) 54 m
(b) 100 m
(c) \( \frac{100}{\sqrt{3}} \) m
(d) \( 100\sqrt{3} \) m
Show Answer & Explanation

Answer: (c) \( \frac{100}{\sqrt{3}} \) m

Explanation:
1. sin 60° = 50/AC, so AC = 50 × 2/√3 = 100/√3 m.

Teacher's Note:
1. The line of sight is the hypotenuse.
2. The vertical gap is 50 m, not 54 m.

Question 37: Raju, 4 m tall, sees a pigeon on top of a 54 m pole. The angle of elevation of the pigeon from his eyes is 60°. The pigeon flies horizontally at constant height and after 8 seconds the angle of elevation from the same point is 45°. (Take √3 = 1.73.) The pigeon is therefore 54 − 4 = 50 m above his eye level. If the distance between the two positions of the pigeon increases, the angle of elevation:
(a) increases
(b) decreases
(c) remains unchanged
(d) can't say
Show Answer & Explanation

Answer: (b) decreases

Explanation:
1. As the pigeon moves farther away at the same height, the horizontal distance grows, so tan(elevation) = 50/distance becomes smaller.
2. The angle decreases.

Teacher's Note:
1. Height is constant; distance increases.
2. A larger denominator gives a smaller ratio.

Question 38: Raju, 4 m tall, sees a pigeon on top of a 54 m pole. The angle of elevation of the pigeon from his eyes is 60°. The pigeon flies horizontally at constant height and after 8 seconds the angle of elevation from the same point is 45°. (Take √3 = 1.73.) The pigeon is therefore 54 − 4 = 50 m above his eye level. The horizontal distance AB between Raju and the pole is:
60°45°ACEBD (a) 50 m
(b) \( \frac{50}{\sqrt{3}} \) m
(c) \( 50\sqrt{3} \) m
(d) \( \frac{100}{\sqrt{3}} \) m
Show Answer & Explanation

Answer: (b) \( \frac{50}{\sqrt{3}} \) m

Explanation:
1. tan 60° = 50/AB, so AB = 50/√3 m.

Teacher's Note:
1. Horizontal distance = height gap / tan 60°.
2. tan 60° = √3.

Question 39: Raju, 4 m tall, sees a pigeon on top of a 54 m pole. The angle of elevation of the pigeon from his eyes is 60°. The pigeon flies horizontally at constant height and after 8 seconds the angle of elevation from the same point is 45°. (Take √3 = 1.73.) The pigeon is therefore 54 − 4 = 50 m above his eye level. How far does the pigeon fly in 8 seconds?
(a) 12.13 m
(b) 19.60 m
(c) 21.1 m
(d) 26.32 m
Show Answer & Explanation

Answer: (c) 21.1 m

Explanation:
1. At 45° the horizontal distance is 50 m. At 60° it was 50/√3 = 50/1.73 = 28.9 m.
2. Distance covered = 50 − 28.9 = 21.1 m.

Teacher's Note:
1. The two horizontal distances from Raju are 28.9 m and 50 m.
2. Subtract the smaller from the larger.

Question 40: Raju, 4 m tall, sees a pigeon on top of a 54 m pole. The angle of elevation of the pigeon from his eyes is 60°. The pigeon flies horizontally at constant height and after 8 seconds the angle of elevation from the same point is 45°. (Take √3 = 1.73.) The pigeon is therefore 54 − 4 = 50 m above his eye level. The speed of the pigeon is (using 21.1 m in 8 s):
(a) 2.64 m/s
(b) 3.88 m/s
(c) 6.7 m/s
(d) 9.3 m/s
Show Answer & Explanation

Answer: (a) 2.64 m/s

Explanation:
1. Speed = distance / time = 21.1/8 = 2.64 m/s (approximately).

Teacher's Note:
1. Speed = distance ÷ time.
2. Use the distance found in the previous step.

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