CBSE Class 10 Mathematics Some Applications of Trigonometry MCQs Set 12

Answer: (d) 30°

Explanation:
1. tan θ = rod/shadow = 1/√3, so θ = 30°.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (b) tan θ = 1/2, so θ ≈ 26.6°

Explanation:
1. tan θ = pole/shadow = 1/2, so θ ≈ 26.57°. (It cannot be 90°, because then the shadow would have zero length.)

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (b) 9√3 cm²

Explanation:
1. The sides are 6 cos 30° = 3√3 cm and 6 sin 30° = 3 cm. Area = 3√3 × 3 = 9√3 cm².

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (a) 10√3 m

Explanation:
1. h cot 30° − h cot 60° = 20, so h(√3 − 1/√3) = 20, i.e. h(2/√3) = 20 and h = 10√3 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (a) √(ab)

Explanation:
1. h = a tan 30° = b tan 60°. Multiplying: h² = ab tan 30° tan 60° = ab, so h = √(ab).

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (a) 720 km/h

Explanation:
1. Horizontal distances from A: 1500√3 cot 60° = 1500 m and 1500√3 cot 30° = 4500 m. Distance flown = 3000 m in 15 s = 200 m/s = 200 × 18/5 = 720 km/h.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (c) about 236.6 m

Explanation:
1. Let the rock be h m high and the horizontal distance x. At the foot of the tower x = h. From the top of the tower (h − 100)/x = tan 30°, so h − 100 = h/√3 and h = 100/(1 − 1/√3) = 100√3/(√3 − 1) ≈ 236.6 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (d) about 21.13 m

Explanation:
1. The tower's foot is at horizontal distance 50 cot 45° = 50 m. The top of the tower is 50 tan 30° = 28.87 m below the cliff top, so the tower is 50 − 28.87 = 21.13 m high.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (c) 150 m

Explanation:
1. Distance between the feet = 50 cot 30° = 50√3 m. Height of hill = 50√3 tan 60° = 50√3 × √3 = 150 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (c) 30°

Explanation:
1. tan θ = 10/(10√3) = 1/√3, so θ = 30°.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (b) 40√3 m

Explanation:
1. sin 60° = 60/L, so L = 60/(√3/2) = 120/√3 = 40√3 m ≈ 69.3 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (a) 30°

Explanation:
1. tan θ = 1/√3, so θ = 30°.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (c) about 4.28 m and 2.14 m

Explanation:
1. The point is 5 − 1.3 = 3.7 m high. Length = 3.7/sin 60° = 3.7 × 2/1.73 ≈ 4.28 m. Distance = 3.7/tan 60° = 3.7/1.73 ≈ 2.14 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (a) 8/√3 m

Explanation:
1. Height of break = 8 tan 30° = 8/√3 m ≈ 4.62 m. (The original height of the tree is 8/√3 + 16/√3 = 8√3 m ≈ 13.86 m.)

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (a) 36.2 m

Explanation:
1. Height above eye level = 20 tan 60° = 20 × 1.73 = 34.6 m. Total = 34.6 + 1.6 = 36.2 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (d) 20(√3 − 1) m

Explanation:
1. The distance is 20 cot 45° = 20 m. Total height = 20 tan 60° = 20√3 m. Tower = 20√3 − 20 = 20(√3 − 1) ≈ 14.64 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (d) 0.8(√3 + 1) m

Explanation:
1. Let the pedestal be h m; the distance is h (45°). Then (h + 1.6)/h = √3, so h = 1.6/(√3 − 1) = 0.8(√3 + 1) ≈ 2.19 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (b) 50/3 m high

Explanation:
1. Distance = 50 cot 60° = 50/√3. Building height = (50/√3) tan 30° = 50/3 m ≈ 16.67 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (b) 43.3 m and 75 m

Explanation:
1. h cot 30° − h cot 60° = 50 gives h(2/√3) = 50, so h = 25√3 ≈ 43.3 m. Distance from A = h cot 30° = 25√3 × √3 = 75 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (b) about 346.4 m

Explanation:
1. The distances from the foot are 150 cot 60° = 86.6 m and 150 cot 30° = 259.8 m. On opposite sides: 86.6 + 259.8 = 346.4 m. (On the same side it would be 259.8 − 86.6 = 173.2 m.)

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (a) 20√3 m

Explanation:
1. h cot 30° − h cot 60° = 40 gives h(2/√3) = 40, so h = 20√3 ≈ 34.64 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (d) 13.7 m

Explanation:
1. h cot 30° − h cot 45° = 10 gives h(√3 − 1) = 10, so h = 10/0.732 ≈ 13.66 ≈ 13.7 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (a) 20√3 m and 20 m

Explanation:
1. Let the height be h. The distances from the pillars are h cot 60° and h cot 30°, so h/√3 + h√3 = 80, giving h(4/√3) = 80 and h = 20√3 m. The point is h/√3 = 20 m from the 60° pillar.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (c) 34.64 m and 40 m

Explanation:
1. The distance is 60 cot 60° = 20√3 ≈ 34.64 m. The top of the post is 34.64 tan 30° = 20 m below the building top, so the post is 60 − 20 = 40 m high.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (c) 4(3 + √3) m

Explanation:
1. Let the height be H and distance d: d = H (45° to the bottom). The top of the small building is 8 m high, so (H − 8)/d = tan 30°, i.e. H − 8 = H/√3. H = 8√3/(√3 − 1) = 4√3(√3 + 1) = 4(3 + √3) ≈ 18.93 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (d) 120(√3 − 1) m ≈ 87.84 m

Explanation:
1. Present height = 120 tan 45° = 120 m. Required height = 120 tan 60° = 120√3 m. Increase = 120(√3 − 1) ≈ 87.84 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (d) 7(√3 + 1) m

Explanation:
1. The distance is 7 cot 45° = 7 m. Above the building top the tower rises 7 tan 60° = 7√3 m. Total = 7 + 7√3 = 7(√3 + 1) ≈ 19.12 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (b) 30 m

Explanation:
1. Distance of P = 10 cot 30° = 10√3 m. Height = 10√3 tan 60° = 30 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (b) 75√3 m

Explanation:
1. h cot 30° − h cot 60° = 150 gives h(2/√3) = 150, so h = 75√3 ≈ 129.9 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.

Answer: (d) 30√2 m

Explanation:
1. The bird's height = 100 sin 30° = 50 m. It is 50 − 20 = 30 m above the girl's level. Distance from girl = 30/sin 45° = 30√2 ≈ 42.4 m.

Teacher's Note:
1. Draw a rough figure in your mind, mark the right triangle(s), and use tan of the given angle to relate height and distance.
2. Angle of elevation and angle of depression are measured from the horizontal line of sight.