CBSE Class 10 Mathematics Some Applications of Trigonometry MCQs Set 13

Mathematics Objective Questions and Answers: 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.

Download Chapter 09 Some Applications of Trigonometry MCQs with Answers

Access the complete set of multiple-choice questions for Chapter 09 Some Applications of Trigonometry below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: A man on the deck of a ship, 8 m above the water, observes the angle of elevation of the top of a hill as 60 degrees and the angle of depression of the base of the hill as 30 degrees. The height of the hill is:
(a) 32 m
(b) 24 m
(c) 8 m
(d) 16 sqrt3 m
Show Answer & Explanation

Answer: (a) 32 m

Explanation:
1. tan 30 = 8/x gives x = 8 sqrt3 m (the distance to the hill). The part above the deck level is x tan 60 = 8 sqrt3 x sqrt3 = 24 m. Height of hill = 24 + 8 = 32 m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 2: A vertical tower is surmounted by a flagstaff of height 5 m. At a point on the plane the angles of elevation of the bottom and the top of the flagstaff are 30 degrees and 60 degrees. The height of the tower is:
(a) 5 m
(b) 5 sqrt3 m
(c) 10 m
(d) 2.5 m
Show Answer & Explanation

Answer: (d) 2.5 m

Explanation:
1. h/x = tan 30 and (h + 5)/x = tan 60. So (h + 5) = 3h, giving h = 2.5 m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 3: From the top of a multistoried building the angles of depression of the top and the bottom of an 8 m tall building are 30 degrees and 45 degrees. The height of the multistoried building is:
(a) \( 4(3-\sqrt3)\ \text{m} \)
(b) \( 8(\sqrt3+1)\ \text{m} \)
(c) \( 4(3+\sqrt3)\ \text{m} \)
(d) \( 8\sqrt3\ \text{m} \)
Show Answer & Explanation

Answer: (c) \( 4(3+\sqrt3)\ \text{m} \)

Explanation:
1. Let the distance between the buildings be x and the taller height h. From the base: h = x tan 45 = x. From the top: h - 8 = x tan 30 = x/sqrt3. So x(1 - 1/sqrt3) = 8, x = 8 sqrt3/(sqrt3 - 1) = 4(3 + sqrt3). The height and the distance are both 4(3 + sqrt3) m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 4: The angle of elevation of an aeroplane from a point on the ground is 45 degrees. After 15 seconds the elevation is 30 degrees. If the aeroplane is flying at a height of 3000 m, its speed is about (use sqrt3 = 1.732):
(a) 438 km/h
(b) 527 km/h
(c) 720 km/h
(d) 380 km/h
Show Answer & Explanation

Answer: (b) 527 km/h

Explanation:
1. Horizontal distances: 3000 cot 45 = 3000 and 3000 cot 30 = 3000 sqrt3. Distance covered = 3000(sqrt3 - 1) = 2196 m in 15 s, that is 146.4 m/s = 527.04 km/h.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 5: A boy flies a kite with 100 m of string at an elevation of 30 degrees. Another boy on a 10 m high roof, standing on the other side, flies his kite at an elevation of 45 degrees. The string length the second boy needs for the kites to meet is:
(a) \( 50\sqrt2\ \text{m} \)
(b) \( 40\sqrt2\ \text{m} \)
(c) \( 40\ \text{m} \)
(d) \( 60\sqrt2\ \text{m} \)
Show Answer & Explanation

Answer: (b) \( 40\sqrt2\ \text{m} \)

Explanation:
1. The first kite is at height 100 sin 30 = 50 m. The second boy's string must reach 50 - 10 = 40 m above the roof at 45 degrees, so x sin 45 = 40 and x = 40 sqrt2 m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 6: The upper part of a vertical tree is broken by the wind and touches the ground, making an angle of 30 degrees with it. If the broken part is 20 m long, the length of the remaining standing part is:
(a) 10 m
(b) 20 m
(c) \( 10\sqrt3\ \text{m} \)
(d) \( 10\sqrt2\ \text{m} \)
Show Answer & Explanation

Answer: (a) 10 m

Explanation:
1. The standing part is the vertical side opposite the 30 degree angle: 20 sin 30 = 10 m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 7: The angle of elevation of the top of a tower from a point on the ground is x. After moving a distance d towards the foot of the tower the angle of elevation is y. The height of the tower is:
(a) \( d(\tan y+\tan x) \)
(b) \( d(\tan y-\tan x) \)
(c) \( \frac{d\tan x\tan y}{\tan y+\tan x} \)
(d) \( \frac{d\tan x\tan y}{\tan y-\tan x} \)
Show Answer & Explanation

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

Explanation:
1. Let the height be h and the final distance be c. h = c tan y and h = (c + d) tan x. So c = h/tan y and h = (h/tan y + d) tan x, giving h(1 - tan x/tan y) = d tan x, hence h = d tan x tan y/(tan y - tan x).

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 8: The angles of elevation of the top of a tower from two points on the same line through its foot, at distances p and q, are complementary. The height of the tower is:
(a) \( pq \)
(b) \( \frac pq \)
(c) \( \sqrt{pq} \)
(d) \( \sqrt{p+q} \)
Show Answer & Explanation

Answer: (c) \( \sqrt{pq} \)

Explanation:
1. If the elevations are A and 90 - A, then h/p = tan A and h/q = cot A. Multiplying, h^2/(pq) = 1, so h = sqrt(pq).

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 9: The angle of elevation of the top of a tower from a point 50 sqrt3 / 3 m from its foot is 60 degrees. The height of the tower is:
(a) 50 sqrt3 m
(b) 20/sqrt3 m
(c) 50 m
(d) -50 m
Show Answer & Explanation

Answer: (c) 50 m

Explanation:
1. Height = distance x tan 60 = (50 sqrt3/3)(sqrt3) = 50 m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 10: The shadow of a tower when the sun's elevation is 30 degrees is 5 m longer than when it is 45 degrees. The height of the tower is:
(a) \( \frac52(\sqrt3-1)\ \text{m} \)
(b) \( \frac52(\sqrt3+1)\ \text{m} \)
(c) \( \frac{5}{\sqrt3+1}\ \text{m} \)
(d) \( 5\sqrt3\ \text{m} \)
Show Answer & Explanation

Answer: (b) \( \frac52(\sqrt3+1)\ \text{m} \)

Explanation:
1. h cot 30 - h cot 45 = 5 gives h(sqrt3 - 1) = 5, so h = 5/(sqrt3 - 1) = (5/2)(sqrt3 + 1) m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 11: From the top of a lighthouse 30 m high the angles of depression of two ships on opposite sides, in line with its foot, are 45 degrees and 30 degrees. The distance between the ships is:
(a) \( 30(1+\sqrt3)\ \text{m} \)
(b) \( 30(\sqrt3-1)\ \text{m} \)
(c) \( 60\ \text{m} \)
(d) \( 30\sqrt3\ \text{m} \)
Show Answer & Explanation

Answer: (a) \( 30(1+\sqrt3)\ \text{m} \)

Explanation:
1. The ships are h cot 45 = 30 m and h cot 30 = 30 sqrt3 m from the foot. Distance between them = 30 + 30 sqrt3 = 30(1 + sqrt3) m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 12: From an aeroplane directly above a straight road the angles of depression of two consecutive kilometre stones on opposite sides of it are 30 degrees and 60 degrees. The height of the aeroplane above the road is:
(a) \( \frac{\sqrt3}{2}\ \text{km} \)
(b) \( \frac14\ \text{km} \)
(c) \( \sqrt3\ \text{km} \)
(d) \( \frac{\sqrt3}{4}\ \text{km} \)
Show Answer & Explanation

Answer: (d) \( \frac{\sqrt3}{4}\ \text{km} \)

Explanation:
1. With height h the stones are h cot 30 and h cot 60 from the foot: h(sqrt3 + 1/sqrt3) = 1, i.e. h (4/sqrt3) = 1, so h = sqrt3/4 km.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 13: A window is 10 m above the ground. The angle of depression of a point P on the ground from the window is 30 degrees and the angle of elevation of the top of the building from P is 60 degrees. The height of the building is:
(a) 10 sqrt3 m
(b) 20 m
(c) 10 m
(d) 30 m
Show Answer & Explanation

Answer: (d) 30 m

Explanation:
1. PQ = 10 cot 30 = 10 sqrt3 m from the foot. Height = PQ tan 60 = 10 sqrt3 x sqrt3 = 30 m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 14: A man on a cliff sees a boat approaching the shore at an angle of depression of 30 degrees. Six minutes later the angle of depression is 60 degrees. The total time taken by the boat from the first position to reach the shore is:
(a) 12 minutes
(b) 6 minutes
(c) 9 minutes
(d) 3 minutes
Show Answer & Explanation

Answer: (c) 9 minutes

Explanation:
1. With cliff height h, the distances are h cot 30 = h sqrt3 and h cot 60 = h/sqrt3. The boat covers h(sqrt3 - 1/sqrt3) = 2h/sqrt3 in 6 min, and the remaining h/sqrt3 takes 3 min. Total 9 minutes.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 15: Two pillars of equal height stand on either side of a 100 m wide road. From a point on the road between them the elevation angles of the tops are 60 degrees and 30 degrees. The point is __ from the pillar with the 60 degree elevation, and each pillar has height:
(a) 75 m from it, 43.3 m high
(b) 25 m from it, 43.3 m high
(c) 50 m from it, 50 m high
(d) 25 m from it, 25 m high
Show Answer & Explanation

Answer: (b) 25 m from it, 43.3 m high

Explanation:
1. Let the point be x from the 60 degree pillar: x tan 60 = (100 - x) tan 30, so 3x = 100 - x and x = 25 m. Height = 25 sqrt3 = 43.3 m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 16: At a point the tangent of the angle of elevation of a tower is 5/12. After walking 240 m nearer the tangent becomes 3/4. The height of the tower is:
(a) 225 m
(b) 180 m
(c) 240 m
(d) 300 m
Show Answer & Explanation

Answer: (a) 225 m

Explanation:
1. h cot(first) - h cot(second) = 240: h(12/5 - 4/3) = 240, so h(16/15) = 240 and h = 225 m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 17: From a window 10 m above the ground in a street the angle of elevation of the top of the opposite house is 60 degrees and the angle of depression of its foot is 45 degrees. The height of the opposite house is:
(a) \( 10(1+\sqrt3)\ \text{m} \)
(b) \( 10(\sqrt3-1)\ \text{m} \)
(c) \( 20\ \text{m} \)
(d) \( 10\sqrt3\ \text{m} \)
Show Answer & Explanation

Answer: (a) \( 10(1+\sqrt3)\ \text{m} \)

Explanation:
1. The distance across the street is 10 cot 45 = 10 m. The part above the window is 10 tan 60 = 10 sqrt3. Height = 10 + 10 sqrt3 = 10(1 + sqrt3) m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 18: A 5 m pole is fixed on a tower. From a point A on the ground the angle of elevation of the top of the pole is 60 degrees and the angle of depression of A from the top of the tower is 45 degrees. The height of the tower is about:
(a) 5 m
(b) 8.66 m
(c) 10 m
(d) 6.83 m
Show Answer & Explanation

Answer: (d) 6.83 m

Explanation:
1. With tower height h and distance x: x = h (45 degrees) and (h + 5)/x = tan 60. So h + 5 = sqrt3 h, h = 5/(sqrt3 - 1) = 2.5(sqrt3 + 1) = 6.83 m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 19: The angle of elevation of a jet from a point A on the ground is 60 degrees. After 15 seconds the elevation is 30 degrees. If the jet flies at 720 km/h, its constant height is about (sqrt3 = 1.732):
(a) 1500 m
(b) 3000 m
(c) 2598 m
(d) 1732 m
Show Answer & Explanation

Answer: (c) 2598 m

Explanation:
1. Distance in 15 s = 200 m/s x 15 = 3000 m. h cot 30 - h cot 60 = 3000, so h(sqrt3 - 1/sqrt3) = h (2/sqrt3) = 3000 and h = 1500 sqrt3 = 2598 m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 20: A tower is 30 m high. The angle of elevation of its top from a point on the ground is 30 degrees. The distance of the point from the foot of the tower is:
(a) \( 10\sqrt3\ \text{m} \)
(b) \( 30\sqrt3\ \text{m} \)
(c) \( 15\ \text{m} \)
(d) \( 60\ \text{m} \)
Show Answer & Explanation

Answer: (b) \( 30\sqrt3\ \text{m} \)

Explanation:
1. distance = height / tan 30 = 30 sqrt3 m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

Question 21: A ladder 10 m long rests against a vertical wall and makes an angle of 60 degrees with the ground. The height it reaches on the wall is:
(a) \( 5\ \text{m} \)
(b) \( 5\sqrt3\ \text{m} \)
(c) \( 10\sqrt3\ \text{m} \)
(d) \( \frac{10}{\sqrt3}\ \text{m} \)
Show Answer & Explanation

Answer: (b) \( 5\sqrt3\ \text{m} \)

Explanation:
1. Height = 10 sin 60 = 10 (sqrt3/2) = 5 sqrt3 m.

Teacher's Note:
1. Angle of elevation: angle above the horizontal when looking up at an object. Angle of depression: angle below the horizontal when looking down. Draw the right triangle, mark the horizontal through the observer, and use tan = height/distance. Alternate angles: the angle of depression from the top equals the angle of elevation of the top from the object.

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

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