CBSE Class 10 Mathematics Some Applications of Trigonometry MCQs Set 08

Answer: (b) 60°

Explanation:
1. Let the shadow be x, so the pole is √3 x.
2. tan θ = √3x / x = √3 = tan 60°, so θ = 60°.

Teacher's Note:
1. The angle of elevation of the Sun is found from tan θ = height / shadow.
2. tan 60° = √3.

Answer: (b) 40 m

Explanation:
1. Let the horizontal distance be d and the common angle x. From the cliff top, the foot of the tower is 20 m below, so tan x = 20/d.
2. The top of the tower is h m above the cliff top, so tan x = h/d. Hence h = 20.
3. Height of tower = 20 + 20 = 40 m.

Teacher's Note:
1. The two angles are equal, so the vertical distances above and below the cliff top are equal.
2. Add the cliff's height to the part above it.

Answer: (d) 4

Explanation:
1. The foot is on the ground, so cos 60° = 2 / (ladder length).
2. 1/2 = 2/L, so L = 4 m.

Teacher's Note:
1. The distance from the wall is the base of the right triangle; the ladder is the hypotenuse.
2. Use cosine for base and hypotenuse.

Answer: (c) \( 75\sqrt{3} \)

Explanation:
1. The angle of depression equals the angle of elevation of the tower top from the car: 30°.
2. tan 30° = 75/x, so 1/√3 = 75/x and x = 75√3 m.

Teacher's Note:
1. Depression from the top equals elevation from the car (alternate angles).
2. tan 30° = height / distance.

Answer: (b) 30°

Explanation:
1. The height is the perpendicular and the string is the hypotenuse.
2. sin θ = 30/60 = 1/2 = sin 30°, so θ = 30°.

Teacher's Note:
1. Use sine for opposite and hypotenuse.
2. sin 30° = 1/2.

Answer: (d) 45°

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

Teacher's Note:
1. Equal height and shadow give tan θ = 1.
2. This is a quick standard result.

Answer: (c) 25

Explanation:
1. tan 45° = H/25, so 1 = H/25 and H = 25 m.

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

Answer: (d) 3 cm

Explanation:
1. BC is opposite ∠A and AC is the hypotenuse.
2. sin 30° = BC/6, so 1/2 = BC/6 and BC = 3 cm.

Teacher's Note:
1. Opposite over hypotenuse is sine.
2. sin 30° = 1/2.

Answer: (a) \( 8\sqrt{3} \) m

Explanation:
1. The vertical difference between the tops is 36 − 24 = 12 m.
2. sin 60° = 12 / (wire length), so wire = 12 / (√3/2) = 24/√3 = 8√3 m.

Teacher's Note:
1. Only the difference in heights matters for the wire's angle.
2. Rationalise 24/√3 to 8√3.

Answer: (c) \( 16\sqrt{3} \) m

Explanation:
1. tan 60° = h/16, so √3 = h/16 and h = 16√3 m.

Teacher's Note:
1. Height = shadow × tan(elevation).
2. tan 60° = √3.

Answer: (b) 25 m

Explanation:
1. The distances from the foot of the point to the banks are 9/tan 45° = 9 m and 9/tan 30° = 9√3 m.
2. Width = 9 + 9√3 = 9 × 2.73 = 24.57 m, which is about 25 m.

Teacher's Note:
1. Use alternate angles: depression equals elevation from each bank.
2. Add the two horizontal distances.

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

Explanation:
1. Let the height be h and the shadow at 60° be x. Then h = x√3.
2. At 30°, h/(x + 50) = 1/√3, so x√3 · √3 = x + 50, i.e. 3x = x + 50 and x = 25.
3. h = 25√3 m.

Teacher's Note:
1. Write the height in terms of the shorter shadow first.
2. Equate the two expressions for h.

Answer: (a) 40 m

Explanation:
1. Let the tower be h m high and the horizontal distance y. From the building top, the foot of the tower gives tan 60° = 60/y, so y = 60/√3.
2. For the tower top, tan 30° = (60 − h)/y, so (60 − h) = y/√3 = 20.
3. Hence h = 40 m.

Teacher's Note:
1. Use the foot of the tower to find the distance, then the top to find the remaining height.
2. Vertical gap above the tower top is 60 − h.

Answer: (c) \( 150(3+\sqrt{3}) \) m

Explanation:
1. Let the nearer ship be x m from the base. Then h = x√3 (60°) and h = x + 300 (45°).
2. x(√3 − 1) = 300, so x = 150(√3 + 1).
3. h = x + 300 = 150(√3 + 1) + 300 = 150(3 + √3) m.

Teacher's Note:
1. The nearer ship has the larger depression angle.
2. Rationalise 300/(√3 − 1) by multiplying by (√3 + 1).

Answer: (d) 45°

Explanation:
1. The height of the tower above the observer's eye is 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's height.
2. Equal vertical and horizontal distances give 45°.

Answer: (a) \( 150\sqrt{3} \)

Explanation:
1. tan 30° = 150/x, so x = 150/tan 30° = 150√3 m.

Teacher's Note:
1. Same method as a standard depression problem.
2. tan 30° = 1/√3.

Answer: (d) 30°, 45°

Explanation:
1. O₁ is vertically above C with ∠O₁AC = 30°, so the angle of depression at O₁ is 30° (alternate angles).
2. O₂ is vertically above B with ∠O₂AB = 45°, so the angle of depression at O₂ is 45°.
3. The answer is 30°, 45°.

Teacher's Note:
1. The angle of depression equals the angle of elevation at the object (alternate angles).
2. Read the angle at A for each position.

Answer: (b) 10 m

Explanation:
1. tan 45° = 10 / shadow, so shadow = 10 m.

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

Answer: (b) \( \sqrt{ab} \)

Explanation:
1. Let the height be h. tan 30° = h/a and tan 60° = h/b.
2. Multiplying: (1/√3)(√3) = h²/(ab), so h² = ab.
3. Hence \( h=\sqrt{ab} \).

Teacher's Note:
1. The farther point has the smaller angle, 30°.
2. Multiplying the two equations cancels the √3.

Answer: (d) 5 m

Explanation:
1. At the same time, the Sun's angle is the same, so height : shadow is constant.
2. 5/2 = 12.5/s, so s = 5 m.

Teacher's Note:
1. Similar triangles or equal tan θ give proportional heights and shadows.
2. 12.5 is 2.5 times 5.

Answer: (c) 2.5 m

Explanation:
1. The girl and the lamp post make similar triangles with the shadow end: tan θ = 1.5/4.5 = 1/3.
2. The lamp post stands at 3 + 4.5 = 7.5 m from the shadow tip, so height = 7.5 × 1/3 = 2.5 m.

Teacher's Note:
1. Total distance from the shadow tip to the post is the sum of 3 m and 4.5 m.
2. Same tan θ for both triangles.

Answer: (a) \( (\sqrt{3}+1)h \) metres

Explanation:
1. Distance from the base to the first ship: h/tan 45° = h.
2. Distance to the second ship: h/tan 30° = √3 h.
3. Ships are on opposite sides, so the distance is h + √3 h = (√3 + 1)h.

Teacher's Note:
1. On opposite sides, distances add; on the same side they subtract.
2. Depression angle equals elevation angle at the ship.

Answer: (c) \( \frac{h}{3} \) m

Explanation:
1. Let the first point be C and the second point D, h m above C. The foot of the tower B is seen from D at 60°, so CB = h/tan 60° = h/√3.
2. From C, the tower height AB = CB tan 30° = (h/√3)(1/√3) = h/3 m.

Teacher's Note:
1. Find the horizontal distance first using the depression angle.
2. Then use the elevation angle at the lower point.

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

Explanation:
1. Let the height be h. At 45° the shadow is h; at 30° the shadow is h√3.
2. h√3 − h = 2x, so h = 2x/(√3 − 1) = (√3 + 1)x.

Teacher's Note:
1. Shadow = h / tan(elevation).
2. Multiply by (√3 + 1) to rationalise.

Answer: (c) \( 16\sqrt{3} \) cm²

Explanation:
1. Longer side = 8 cos 30° = 4√3 cm and shorter side = 8 sin 30° = 4 cm.
2. Area = 4√3 × 4 = 16√3 cm².

Teacher's Note:
1. The diagonal is the hypotenuse of the right triangle formed by two adjacent sides.
2. The side next to the 30° angle is the longer one.

Answer: (a) 60°

Explanation:
1. cos θ = 0.5 = cos 60°, so θ = 60°.

Teacher's Note:
1. Recall cos 60° = 1/2.
2. The ladder's inclination to the ground is the angle of elevation.

Answer: (c) 3.7 m

Explanation:
1. BD = AD − AB = 5 − 1.3 = 3.7 m.

Teacher's Note:
1. BD is the height of the repair point above the ground.
2. Subtract the 1.3 m gap from the pole's height.

Answer: (d) 4.28 m

Explanation:
1. sin 60° = BD/BC, so BC = 3.7/(√3/2) = 7.4/1.73 = 4.28 m.

Teacher's Note:
1. The ladder is the hypotenuse BC.
2. Use sine because BD is opposite the 60° angle.

Answer: (b) 2.14 m

Explanation:
1. tan 60° = BD/DC, so DC = 3.7/1.73 = 2.14 m.

Teacher's Note:
1. DC is the base of the right triangle.
2. Use tan 60° = √3.

Answer: (c) 8.56 m

Explanation:
1. The new BD is 7.4 m. Ladder = 7.4/sin 60° = 14.8/1.73 = 8.56 m.

Teacher's Note:
1. Doubling the triangle's sides doubles the ladder: 2 × 4.28.
2. The angle does not change.

Answer: (a) \( 12\sqrt{3} \) m

Explanation:
1. BC = OC tan 30° = 36/√3 = 12√3 m.

Teacher's Note:
1. Section B is the lower part, up to the 30° line of sight.
2. 36/√3 = 12√3.

Answer: (c) \( 12(3-\sqrt{3}) \) m

Explanation:
1. AC = OC tan 45° = 36 m.
2. Section A = AC − BC = 36 − 12√3 = 12(3 − √3) m.

Teacher's Note:
1. The 45° line gives the full tower height.
2. Subtract section B from the total.

Answer: (d) \( 36\sqrt{2} \) m

Explanation:
1. cos 45° = OC/OA, so OA = 36/(1/√2) = 36√2 m.

Teacher's Note:
1. The wire is the hypotenuse.
2. Use cosine for adjacent and hypotenuse.

Answer: (c) \( 24\sqrt{3} \) m

Explanation:
1. cos 30° = OC/OB, so OB = 36/(√3/2) = 72/√3 = 24√3 m.

Teacher's Note:
1. Rationalise 72/√3 to 24√3.
2. cos 30° = √3/2.

Answer: (b) 45°

Explanation:
1. The angle of elevation of the top from O is 45°, and the angle of depression from the top to O is equal to it (alternate angles).
2. So the answer is 45°.

Teacher's Note:
1. Elevation and depression for the same line of sight are equal.
2. The top of the tower is the top of section A.

Answer: (a) 60°

Explanation:
1. tan θ = 9/(3√3) = √3 = tan 60°, so θ = 60°.

Teacher's Note:
1. Opposite is 9 m (height) and adjacent is 3√3 m.
2. tan 60° = √3.

Answer: (c) 1.038 m

Explanation:
1. At a height of 1.8 m the ladder is at a horizontal distance x from its foot where tan 60° = 1.8/x.
2. x = 1.8/1.73 = 1.04 m (1.038 m).

Teacher's Note:
1. The person's head is 1.8 m above the floor.
2. Use tan 60° = height/horizontal distance.

Answer: (b) 10.38 m

Explanation:
1. sin 60° = 9/L, so L = 9 × 2/√3 = 18/1.73 = 10.38 m.

Teacher's Note:
1. The ladder is the hypotenuse of the right triangle.
2. sin 60° = √3/2.

Answer: (c) 30°

Explanation:
1. The angle with the floor is 60° and the wall is perpendicular to the floor, so the angle with the wall = 90° − 60° = 30°.

Teacher's Note:
1. Angles in a right triangle add to 180°.
2. The two acute angles are complementary.

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

Explanation:
1. The Reason is the definition of tan θ, so it is true.
2. tan 30° = AB/20, so AB = 20/√3 = 20/1.73 = 11.56 m. The Assertion is true and follows from the Reason.

Teacher's Note:
1. AB is opposite to ∠C and BC is adjacent to it.
2. Use tan when the opposite and adjacent sides are involved.