CBSE Class 10 Mathematics Some Applications of Trigonometry MCQs Set 06

Answer: (b) \( 25(\sqrt{3} + 1) \) m

Explanation:
1. Let the height be h. The nearer car is at distance h (angle 45°) and the farther car is at distance \( h\sqrt{3} \) (angle 30°).
2. \( h\sqrt{3} - h = 50 \Rightarrow h = \frac{50}{\sqrt{3} - 1} = \frac{50(\sqrt{3} + 1)}{2} = 25(\sqrt{3} + 1) \) m.

Teacher's Note:
1. Write each distance as h cot (angle) and subtract.
2. Rationalise the denominator.

Answer: (d) 415.68 km/h

Explanation:
1. Horizontal distances: at 60°: \( \frac{1}{\tan 60°} = \frac{1}{\sqrt{3}} \) km; at 30°: \( \frac{1}{\tan 30°} = \sqrt{3} \) km.
2. Distance covered = \( \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} = 1.1547 \) km in 10 s.
3. Speed = 1.1547 × 360 = 415.68 km/h.

Teacher's Note:
1. The height stays constant, so only horizontal distances change.
2. 1 km in 10 s is 360 km/h per km.

Answer: (a) 94.79 m

Explanation:
1. Let the height be h. At 45° the distance is h and at 60° the distance is \( \frac{h}{\sqrt{3}} \).
2. \( h - \frac{h}{\sqrt{3}} = 40 \Rightarrow h = \frac{40}{1 - 1/1.73} = \frac{40}{0.42197} \approx 94.79 \) m.

Teacher's Note:
1. Write both distances in terms of h and subtract.
2. Use 1/1.73 ≈ 0.578.

Answer: (a) \( 5(3 + \sqrt{3}) \) m

Explanation:
1. Distance between the building and the tower = 15 cot 45° = 15 m (from the depression of the foot).
2. Tower height = 15 + 15 tan 30° = \( 15 + 5\sqrt{3} = 5(3 + \sqrt{3}) \) m.

Teacher's Note:
1. The depression of the foot gives the horizontal distance.
2. Add the height of the building to the part above its top.

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

Explanation:
1. The vertical difference between the two tops is 30 − 24 = 6 m.
2. \( \sin 60° = \frac{6}{L} \Rightarrow L = \frac{6 \times 2}{\sqrt{3}} = 4\sqrt{3} \) m.

Teacher's Note:
1. Use the difference in heights as the opposite side.
2. sin 60° = √3/2.

Answer: (a) 15 m

Explanation:
1. tan 45° = height/15 and tan 45° = 1, so the height = 15 m.

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

Answer: (a) 45°

Explanation:
1. tan θ = height/shadow = 1, so θ = 45°.

Teacher's Note:
1. Equal opposite and adjacent sides give tan θ = 1.
2. An angle of elevation is always less than 90°.

Answer: (a) 60°

Explanation:
1. \( \tan\theta = \frac{6}{2\sqrt{3}} = \sqrt{3} \), so θ = 60°.

Teacher's Note:
1. tan θ = height/shadow.
2. tan 60° = √3.

Answer: (b) \( 30\sqrt{3} \)

Explanation:
1. tan 60° = h/30, so \( h = 30\sqrt{3} \) m.

Teacher's Note:
1. The side opposite to the angle is the height.
2. tan 60° = √3.

Answer: (a) 20 m

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

Teacher's Note:
1. tan 45° = 1.
2. The height is equal to the distance.

Answer: (a) 5 sec

Explanation:
1. Let the tower height be h. Distances: at 30°, \( h\sqrt{3} \); at 60°, \( \frac{h}{\sqrt{3}} \).
2. In 10 s the car covers \( h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{2h}{\sqrt{3}} \), so the speed is \( \frac{2h}{10\sqrt{3}} \).
3. Time for the remaining \( \frac{h}{\sqrt{3}} \) = \( \frac{h/\sqrt{3}}{2h/(10\sqrt{3})} = 5 \) s.

Teacher's Note:
1. Find the distance covered in 10 s and the distance left.
2. The ratio of the two distances is 2 : 1.

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

Explanation:
1. The foot of the ladder is opposite to the angle of 60° at the top between the ladder and the wall.
2. Distance = 20 sin 60° = \( 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \) m.

Teacher's Note:
1. The ladder is the hypotenuse and the angle is with the wall.
2. The side opposite the angle is 20 sin 60°.

Answer: (b) 38 m

Explanation:
1. Height above eye level = 20 tan 60° = 20 × 1.73 = 34.6 m.
2. Chimney height = 3 + 34.6 = 37.6 m ≈ 38 m.

Teacher's Note:
1. Add the observer's height to the height above eye level.
2. Round at the end.

Answer: (a) 74.64 m

Explanation:
1. The broken part is the hypotenuse: \( \cos 60° = \frac{20}{L} \Rightarrow L = 40 \) m.
2. The standing part = 20 tan 60° = 20 × 1.732 = 34.64 m.
3. Total height = 40 + 34.64 = 74.64 m.

Teacher's Note:
1. The tree height is the standing part plus the broken part.
2. Do not forget to add the broken length.

Answer: (a) 7.5 m

Explanation:
1. sin 30° = height/15, so the height = 15 × \( \frac{1}{2} \) = 7.5 m.

Teacher's Note:
1. The rope is the hypotenuse.
2. sin 30° = 1/2.

Answer: (c) 60°

Explanation:
1. In right triangle ABC, \( \tan\angle ACB = \frac{AB}{BC} = \frac{5\sqrt{3}}{5} = \sqrt{3} \), so ∠ACB = 60°.
2. The angle of depression from A equals ∠ACB (alternate angles with the horizontal AD).

Teacher's Note:
1. Alternate angles are equal, so the angle of depression equals the angle of elevation from C.
2. tan 60° = √3.

Answer: (a) \( \frac{8}{3} \) m (about 2.67 m)

Explanation:
1. The tip of the shadow is 3.2 + 4.8 = 8 m from the lamp-post. The girl and the lamp-post form similar right triangles with the shadow.
2. \( \frac{H}{8} = \frac{1.6}{4.8} \Rightarrow H = \frac{8}{3} \approx 2.67 \) m.

Teacher's Note:
1. Use the full base 8 m for the big triangle.
2. The ratio height/base is the same for both triangles.

Answer: (a) 30 m

Explanation:
1. Distance of P from the building = 10 cot 30° = \( 10\sqrt{3} \) m.
2. Height of the building = \( 10\sqrt{3} \times \tan 60° = 10\sqrt{3} \times \sqrt{3} = 30 \) m.

Teacher's Note:
1. The depression from the window gives the horizontal distance.
2. The same distance is used for the elevation of the top.

Answer: (a) 30°

Explanation:
1. The height = x tan 45° = x. The new shadow = x + (√3 − 1)x = √3 x.
2. \( \tan\theta = \frac{x}{\sqrt{3}x} = \frac{1}{\sqrt{3}} \), so θ = 30°.

Teacher's Note:
1. The height does not change.
2. Use the new shadow length.

Answer: (a) about 64.95 m and 37.5 m

Explanation:
1. Let the point be d m from the pillar seen at 60°. Then h = d tan 60° = \( d\sqrt{3} \) and h = (150 − d) tan 30° = \( \frac{150 - d}{\sqrt{3}} \).
2. 3d = 150 − d, so d = 37.5 m and h = 37.5√3 ≈ 64.95 m.

Teacher's Note:
1. Equal heights give an equation in d.
2. The point is nearer the pillar with the larger angle.

Answer: (a) 154 m

Explanation:
1. tan θ = 4/3 gives sin θ = 4/5.
2. Height above the roof = 180 × 4/5 = 144 m, so the height from the ground = 144 + 10 = 154 m.

Teacher's Note:
1. Find sin θ from tan θ using the 3-4-5 triangle.
2. Add the height of the building.

Answer: (a) 5.22 m

Explanation:
1. RB = 5 − 1.3 = 3.7 m. In the right triangle RBL with ∠L = 45°: \( \sin 45° = \frac{RB}{RL} \).
2. RL = 3.7 × √2 = 3.7 × 1.41 = 5.217 ≈ 5.22 m.

Teacher's Note:
1. The ladder is the hypotenuse of triangle RBL.
2. RB is the vertical height reached.

Answer: (a) 189.28 m

Explanation:
1. Distances from the foot: 120 cot 60° = \( 40\sqrt{3} \) = 69.28 m and 120 cot 45° = 120 m.
2. The cars are on opposite sides, so the distance between them = 69.28 + 120 = 189.28 m.

Teacher's Note:
1. Opposite sides means the two distances add.
2. cot 60° = 1/√3.

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

Explanation:
1. Let the point be d from the 60° pole: \( d\sqrt{3} = \frac{80 - d}{\sqrt{3}} \), so 3d = 80 − d and d = 20 m.
2. Height = 20 tan 60° = \( 20\sqrt{3} \) m (about 34.64 m); the point is 20 m and 60 m from the poles.

Teacher's Note:
1. The same height is written from both sides.
2. Solve for d first.

Answer: (a) 7.5 m

Explanation:
1. The height is the side adjacent to the angle with the wall: height = 15 cos 60° = 15 × 1/2 = 7.5 m.

Teacher's Note:
1. The angle is measured from the wall, so use cos.
2. cos 60° = 1/2.

Answer: (a) \( \sqrt{2} \) cm

Explanation:
1. In the first triangle, sin 45° = 2/hypotenuse, so the diagonal = \( 2\sqrt{2} \) cm.
2. In the second triangle, cos 60° = '?'/diagonal, so '?' = \( 2\sqrt{2} \times \frac{1}{2} = \sqrt{2} \) cm.

Teacher's Note:
1. The diagonal is the hypotenuse of both triangles.
2. The side next to 60° uses cos 60°.

Answer: (b) 45°

Explanation:
1. Height of the tower above the eye = 30 − 1.5 = 28.5 m.
2. tan θ = 28.5/28.5 = 1, so θ = 45°.

Teacher's Note:
1. Subtract the observer's height from the tower height.
2. Equal sides give tan θ = 1.

Answer: (a) 23.07 m

Explanation:
1. First pole: 20 tan 60° = 20√3 m. Second pole: 20 tan 30° = \( \frac{20}{\sqrt{3}} \) m.
2. Difference = \( 20\sqrt{3} - \frac{20}{\sqrt{3}} = \frac{40}{\sqrt{3}} = \frac{40 \times 1.73}{3} = 23.07 \) m.

Teacher's Note:
1. Find each pole's height separately.
2. Take the 20 m as the adjacent side in both cases.

Answer: (a) 23.66 m

Explanation:
1. Width of the street = 15 cot 45° = 15 m.
2. Part above the window = 15 tan 30° = \( 5\sqrt{3} \) = 8.66 m.
3. House height = 15 + 8.66 = 23.66 m.

Teacher's Note:
1. The foot's depression gives the street width.
2. Add the 15 m below the window level.

Answer: (a) 21.5 m

Explanation:
1. Height above eye level = \( 20\sqrt{3} \times \tan 30° = 20 \) m.
2. Chimney height = 20 + 1.5 = 21.5 m.

Teacher's Note:
1. Add the observer's height at the end.
2. √3 × (1/√3) = 1.

Answer: (a) 173 m

Explanation:
1. Distance of the first man: 75 cot 30° = 75√3 = 129.75 m. Second man: 75 cot 60° = 25√3 = 43.25 m.
2. They are on opposite sides, so the distance = 129.75 + 43.25 = 173 m (that is, 100√3).

Teacher's Note:
1. Opposite sides means add the two distances.
2. 75√3 + 25√3 = 100√3.

Answer: (a) about 63.09 m

Explanation:
1. Distance of the ship from the lighthouse = 40 cot 45° = 40 m.
2. The ship's top is above the lighthouse top by 40 tan 30° = 23.09 m.
3. Height of the ship = 40 + 23.09 = 63.09 m, that is \( \frac{40(3 + \sqrt{3})}{3} \) m.

Teacher's Note:
1. The foot's depression gives the distance.
2. The ship is taller than the lighthouse.

Answer: (a) \( \frac{200}{\sqrt{3}} \) m (about 115.5 m)

Explanation:
1. tan 30° = altitude/200, so altitude = \( 200 \times \frac{1}{\sqrt{3}} \approx 115.47 \) m.

Teacher's Note:
1. The landmark is the foot of the vertical line below the drone.
2. tan 30° = 1/√3.

Answer: (c) A is true, but R is false.

Explanation:
1. tan θ = height/distance. If both increase by 20% the ratio stays the same, so the angle is unchanged and A is true.
2. tan θ is perpendicular/base, not base/perpendicular, so R is false.

Teacher's Note:
1. A ratio does not change when both terms are multiplied by the same factor.
2. tan θ is the ratio of the perpendicular to the base.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. sin 30° = 10/L, so L = 20 m and A is true.
2. R is the definition used to get this, so R explains A.

Teacher's Note:
1. The string is the hypotenuse.
2. Both statements are true and linked.

Answer: (d) A is false, but R is true.

Explanation:
1. The height is adjacent to the 60° angle with the wall: 20 cos 60° = 10 m, so A is false.
2. R is the correct definition of cosine, so R is true.

Teacher's Note:
1. Check which side is adjacent to the given angle.
2. A is false because the height is 10 m.

Answer: (c) A is true, but R is false.

Explanation:
1. Height = 10 sin 30° = 5 m, so A is true.
2. sin θ = perpendicular/hypotenuse, not hypotenuse/perpendicular, so R is false.

Teacher's Note:
1. Evaluate A with the correct definition.
2. R is the reciprocal of the definition.

Answer: (d) A is false, but R is true.

Explanation:
1. Equal shadow and height give tan θ = 1, so θ = 45°, not 60°; A is false.
2. R is the Pythagoras theorem and is true.

Teacher's Note:
1. Equal legs mean 45°.
2. R is a true statement about right triangles.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. height = distance × tan(angle of elevation), so A is true.
2. R is the definition of tangent that makes this possible, so R explains A.

Teacher's Note:
1. The height is the side opposite the angle.
2. The distance is the adjacent side.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A is true: heights and distances problems are solved with trigonometric ratios.
2. R is true and explains why: the ratios link angles with side lengths.

Teacher's Note:
1. R gives the reason for the usefulness stated in A.
2. Both statements are true.