CBSE Class 10 Mathematics Some Applications of Trigonometry MCQs Set 15

Answer: (a) \( \frac{d tan x tan y}{tan y − tan x} \)

Explanation:
1. Let the height be h. Then the initial distance is h cot x and the new distance is h cot y.
2. d = h cot x − h cot y = h(cot x − cot y), so h = d/(cot x − cot y).
3. Writing cot as 1/tan: h = d tan x tan y/(tan y − tan x).

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (c) \( \frac{5}{2} \)(√3 + 1)

Explanation:
1. Let the height be h. Shadows: h cot 30° = h√3 and h cot 45° = h.
2. h√3 − h = 5, so h = 5/(√3 − 1) = 5(√3 + 1)/2.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (c) 15√3 m

Explanation:
1. Let the height be h. h cot 30° − h cot 60° = 30 gives h(√3 − 1/√3) = 30.
2. h × 2/√3 = 30, so h = 15√3 m.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (a) 9 minutes

Explanation:
1. Let the height be h. Distance at 30° = h√3 and at 60° = h/√3.
2. Distance covered in 18 minutes = h√3 − h/√3 = 2h/√3.
3. The remaining distance h/√3 is half of this, so it takes 9 minutes.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (b) 60°

Explanation:
1. Angle with the wall θ satisfies tan θ = (distance of foot from wall)/(height reached) = 20√3/20 = √3.
2. So θ = 60°.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (d) 15 sec 30° m

Explanation:
1. The broken part is the hypotenuse of a right triangle with base 15 m and angle 30°.
2. cos 30° = 15/hypotenuse, so the hypotenuse = 15 sec 30° (= 10√3 m).

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (c) 10√3 m

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

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (c) 200

Explanation:
1. Distance of the first ship from the foot = 100 cot 30° = 100√3 and of the second = 100 cot 45° = 100.
2. The directions are at right angles, so the distance between ships = √(100² × 3 + 100²) = 200 m.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (a) 30°

Explanation:
1. tan θ = height/shadow = 1/√3.
2. So θ = 30°.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (c) 10√3 m

Explanation:
1. Shadow at 30° = 15 cot 30° = 15√3 m. Shadow at 60° = 15 cot 60° = 15/√3 = 5√3 m.
2. Difference = 15√3 − 5√3 = 10√3 m.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (b) 120(√3 − 1) m

Explanation:
1. Present height = 120 tan 45° = 120 m.
2. Required height = 120 tan 60° = 120√3 m.
3. Extra height = 120√3 − 120 = 120(√3 − 1) m.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (a) 100/3 m

Explanation:
1. Let the distance from the midpoint to each foot be d. First pole: 100 = d tan 60° = d√3, so d = 100/√3.
2. Second pole: h = d tan 30° = (100/√3)(1/√3) = 100/3 m.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (a) \( \frac{√3 + 1}{4} \) km

Explanation:
1. Let the height be h. h cot 30° − h cot 75° = 1.
2. cot 75° = 2 − √3, so h(√3 − 2 + √3) = 1, giving h = 1/(2√3 − 2) = (√3 + 1)/4 km.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (b) 50 m

Explanation:
1. Height = distance × tan 60° = (50√3/3) × √3 = 50 m.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (a) 2598 m

Explanation:
1. Speed = 720 km/h = 200 m/s, so distance flown in 15 s = 3000 m.
2. Let the height be h: h cot 30° − h cot 60° = 3000, so h(√3 − 1/√3) = 3000.
3. h × 2/√3 = 3000 gives h = 1500√3 = 2598 m.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (b) 8 m

Explanation:
1. Let the height be h. tan θ = h/4 and tan(90° − θ) = h/16.
2. Since tan θ × cot θ = 1: (h/4)(h/16) = 1, so h² = 64 and h = 8 m.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (c) 40 m

Explanation:
1. The horizontal distance to the hill = 10 cot 30° = 10√3 m.
2. The part of the hill above the observer = 10√3 × tan 60° = 30 m.
3. Height of the hill = 10 + 30 = 40 m.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (d) 200√3 m

Explanation:
1. Height = 200 tan 60° = 200√3 m.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (c) 18 m

Explanation:
1. The broken part is the hypotenuse of a right triangle with legs 5 and 12, so it is 13 m.
2. Original height = 5 + 13 = 18 m.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.

Answer: (d) 12 m

Explanation:
1. For complementary angles with points at distances a and b: h² = ab.
2. h² = 9 × 16 = 144, so h = 12 m.

Teacher's Note:
1. Draw a neat figure and mark the right triangles formed by the angles of elevation or depression.
2. Use tan θ = height/base and remember that the angle of depression equals the angle of elevation from the ground.