CBSE Class 10 Mathematics Some Applications of Trigonometry MCQs Set 14

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 09 Some Applications of Trigonometry

Explore reliable objective questions for Chapter 09 Some Applications of Trigonometry tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

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

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: The angle of elevation of a ladder leaning against a wall is 60 degrees and the foot of the ladder is 9.5 m away from the wall. The length of the ladder is
(a) 9.5 m
(b) 16.45 m
(c) 19 m
(d) 11 m
Show Answer & Explanation

Answer: (c) 19 m

Explanation:
1. Let the ladder be \( x \) metres: \( \cos 60 \) degrees = \( 9.5/x \).
2. \( x = 9.5 / (1/2) = 19 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 2: If the length of the shadow cast by a pole is \(\sqrt3\) times the length of the pole, the angle of elevation of the sun is
(a) 60 degrees
(b) 30 degrees
(c) 45 degrees
(d) 15 degrees
Show Answer & Explanation

Answer: (b) 30 degrees

Explanation:
1. tan(angle) = pole / shadow = \( 1/ \sqrt3 \).
2. So the angle of elevation is 30 degrees.

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 3: The angle of elevation of the top of a hill from the foot of a tower is 60 degrees and the angle of elevation of the top of the tower from the foot of the hill is 30 degrees. If the tower is 50 m high, the height of the hill is
(a) 150 m
(b) 100 m
(c) 50 m
(d) 86.6 m
Show Answer & Explanation

Answer: (a) 150 m

Explanation:
1. Distance between the feet \( x = 50 \cot 30 \) degrees = \( 50 \sqrt3 \).
2. Height of hill = \( x \tan 60 \) degrees = \( 50 \sqrt3 x \sqrt3 = 150 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 4: An observer is on a tree 8 m high. The angles of depression of the feet of two poles, one on each bank of a river directly opposite and in line with the foot of the tree, are 60 degrees and 30 degrees. The width of the river is
(a) 13.86 m
(b) 18.475 m
(c) 8 m
(d) 24 m
Show Answer & Explanation

Answer: (b) 18.475 m

Explanation:
1. Distances from the foot of the tree: \( 8 \cot 60 = 8/ \sqrt3 \) and \( 8 \cot 30 = 8 \sqrt3 \).
2. Width = \( 8/ \sqrt3 + 8 \sqrt3 = 32/ \sqrt3 = 18.475 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 5: The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60 degrees. At a point Y, 40 m vertically above X, the angle of elevation is 45 degrees. The height of the tower is
(a) 94.64 m
(b) 54.64 m
(c) 40 m
(d) 109.3 m
Show Answer & Explanation

Answer: (a) 94.64 m

Explanation:
1. Let the horizontal distance be \( x \): from \( Y \), the tower rises \( x \) above Y (45 degrees), so PQ = \( x + 40 \).
2. From \( X \): \( (x + 40)/x = \sqrt3 \), so x = 40/(\(\sqrt3\) - 1) = 54.64 m.
3. Height PQ = \( 54.64 + 40 = 94.64 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 6: Two pillars of equal height stand on either side of a road 100 m wide. At a point on the road between them the angles of elevation of the tops are 60 degrees and 30 degrees. The distance of the point from the nearer pillar and the height of each pillar are
(a) 50 m and 86.6 m
(b) 75 m and 43.3 m
(c) 25 m and 86.6 m
(d) 25 m and 43.3 m
Show Answer & Explanation

Answer: (d) 25 m and 43.3 m

Explanation:
1. Let the point be \( x \) from the first pillar: \( h = x \tan 60 = \sqrt3 x \) and \( h = (100 - x) \tan 30 = (100 - x)/ \sqrt3 \).
2. \( 3x = 100 - x \), so \( x = 25 \) and \( h = 25 x 1.732 = 43.3 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 7: From the top of a building 60 m high the angles of depression of the top and the bottom of a tower are 30 degrees and 60 degrees. The height of the tower is
(a) 40 m
(b) 20 m
(c) 30 m
(d) 60 m
Show Answer & Explanation

Answer: (a) 40 m

Explanation:
1. Distance between building and tower = \( 60 \cot 60 = 20 \sqrt3 \).
2. The top of the tower is seen at 30 degrees, so drop below the building top = \( 20 \sqrt3 \tan 30 = 20 m \).
3. Height of tower = \( 60 - 20 = 40 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 8: A man standing on the deck of a ship, 10 m above water level, 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 distance of the hill from the ship and the height of the hill are
(a) 10 m and 30 m
(b) 17.32 m and 30 m
(c) 17.32 m and 40 m
(d) 30 m and 40 m
Show Answer & Explanation

Answer: (c) 17.32 m and 40 m

Explanation:
1. Distance = \( 10 \cot 30 = 10 \sqrt3 = 17.32 m \).
2. Height above deck level = \( 10 \sqrt3 \tan 60 = 30 m \), so the hill height is \( 10 + 30 = 40 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 9: The angle of depression of a boat from a bridge 25 m high is 45 degrees. The horizontal distance of the boat from the bridge is
(a) 12.5 m
(b) 25 m
(c) 43.3 m
(d) 35.4 m
Show Answer & Explanation

Answer: (b) 25 m

Explanation:
1. Distance = \( 25 \cot 45 \) degrees = \( 25 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 10: The slope of a hill makes an angle of 60 degrees with the horizontal. If one has to walk 500 m to reach the top, the height of the hill is
(a) \(250\sqrt3\text{ m}\)
(b) \(250\text{ m}\)
(c) \(500\sqrt3\text{ m}\)
(d) \(\dfrac{500}{\sqrt3}\text{ m}\)
Show Answer & Explanation

Answer: (a) \(250\sqrt3\text{ m}\)

Explanation:
1. Height = \( 500 \sin 60 \) degrees = \( 500 x \) root \( 3 / 2 = 250 \) root \( 3 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 11: A tree is broken by the wind. The top touches the ground at an angle of 30 degrees at a distance of 30 m from the root. The whole height of the tree is
(a) 57.96 m
(b) \(30\sqrt3\text{ m}\approx51.96\text{ m}\)
(c) 60 m
(d) 34.64 m
Show Answer & Explanation

Answer: (b) \(30\sqrt3\text{ m}\approx51.96\text{ m}\)

Explanation:
1. Remaining standing part = \( 30 \tan 30 = 10 \) root \( 3 = 17.32 m \).
2. Broken part = \( 30 / \cos 30 = 20 \) root \( 3 = 34.64 m \).
3. Total = 30 root \( 3 = 51.96 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 12: The angle of elevation of the top of a tower at a distance of 150 m from its foot on a horizontal plane is 60 degrees. The height of the tower is
(a) 86.6 m
(b) 150 m
(c) 300 m
(d) 259.8 m
Show Answer & Explanation

Answer: (d) 259.8 m

Explanation:
1. Height = \( 150 \tan 60 \) degrees = \( 150 x 1.732 = 259.8 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 13: The angle of elevation of the top of a tower from two points at distances p and q from its base, on the same straight line with it, are complementary. The height of the tower is
(a) \(\sqrt{pq}\)
(b) \(pq\)
(c) \(\dfrac{p+q}{2}\)
(d) \(\sqrt{p^2+q^2}\)
Show Answer & Explanation

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

Explanation:
1. With height \( h \): \( \tan A = h/p \) and \( \tan(90 - A) = \cot A = h/q \).
2. Multiplying: \( 1 = h^2/(pq) \), so \( h \) = square root of pq.

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 14: A tower stands vertically on the ground. From the top and the foot of a building of height a, the angles of elevation of the top of the tower are 30 degrees and 45 degrees. If the tower and the building stand at the same level, the height of the tower is
(a) \(a(3+\sqrt3)/2\)
(b) \(a\sqrt3\)
(c) \(a(\sqrt3-1)\)
(d) \(a(\sqrt3+1)\)
Show Answer & Explanation

Answer: (a) \(a(3+\sqrt3)/2\)

Explanation:
1. Let the tower height be \( h \) and the distance \( d \). From the foot: \( h = d \tan 45 = d \).
2. From the top of the building: \( h - a = d \tan 30 \) = d/root 3.
3. h(1 - \( 1/ \sqrt3 \)) = a, so \( h = a \sqrt3 / \)(\( \sqrt3 - 1) = a(3 + \sqrt3 \))/2.

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 15: On level ground the angle of elevation of the top of a tower is 30 degrees. On moving 20 m nearer, the angle of elevation is 60 degrees. The height of the tower is
(a) \(10\sqrt3\text{ m}\)
(b) \(20\sqrt3\text{ m}\)
(c) \(10(\sqrt3-1)\text{ m}\)
(d) \(20\text{ m}\)
Show Answer & Explanation

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

Explanation:
1. \( h \cot 30 - h \cot 60 = 20 \).
2. h(\( \sqrt3 - 1/ \sqrt3 \)) = 20, so h(\( 2/ \sqrt3 \)) = 20 and \( h = 10 \sqrt3 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 16: A pole stands vertically inside a triangular park ABC. If the angle of elevation of the top of the pole from each corner of the park is the same, then in triangle ABC the foot of the pole is at the
(a) centroid
(b) incentre
(c) circumcentre
(d) orthocentre
Show Answer & Explanation

Answer: (c) circumcentre

Explanation:
1. Equal angles of elevation mean the foot is at equal distances from \( A, B \) and \( C \).
2. The point equidistant from all three vertices is the circumcentre.

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 17: A bridge across a river makes an angle of 45 degrees with the river bank. If the length of the bridge across the river is 150 m, the width of the river is
(a) \(150\text{ m}\)
(b) \(75\text{ m}\)
(c) \(75\sqrt2\text{ m}\)
(d) \(150\sqrt2\text{ m}\)
Show Answer & Explanation

Answer: (c) \(75\sqrt2\text{ m}\)

Explanation:
1. Width = \( 150 \sin 45 \) degrees = 150/root \( 2 = 75 \) root \( 2 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 18: The angle of elevation of the top of a tower from a point on the ground is 30 degrees. After walking 30 m towards the tower the angle of elevation becomes 60 degrees. The height of the tower is
(a) \(15\sqrt3\text{ m}\)
(b) \(30\sqrt3\text{ m}\)
(c) \(15\text{ m}\)
(d) \(10\sqrt3\text{ m}\)
Show Answer & Explanation

Answer: (a) \(15\sqrt3\text{ m}\)

Explanation:
1. h(\( \sqrt3 - 1/ \sqrt3 \)) = 30, so \( h = 15 \sqrt3 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 19: The angle of depression of the top and the bottom of a tower as seen from the top of a 100 m high cliff are 30 degrees and 60 degrees. The height of the tower is
(a) 66.67 m
(b) 33.33 m
(c) 50 m
(d) 100 m
Show Answer & Explanation

Answer: (a) 66.67 m

Explanation:
1. Distance = \( 100 \cot 60 = 100/ \sqrt3 \).
2. Drop to the tower top (= \( 100/ \sqrt3 \)) tan 30 = 33.33 m.
3. Tower height = \( 100 - 33.33 = 66.67 m \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 20: The angle of elevation of an aeroplane from a point P on the ground is 60 degrees. After a flight of 15 seconds the angle of elevation becomes 30 degrees. If the aeroplane is flying at a constant height of 1500 root 3 m, its speed is
(a) 200 km/h
(b) 360 km/h
(c) 1000 km/h
(d) 720 km/h
Show Answer & Explanation

Answer: (d) 720 km/h

Explanation:
1. Horizontal distances: \( 1500 \sqrt3 \cot 60 = 1500 m \) and \( 1500 \sqrt3 \cot 30 = 4500 m \).
2. Distance flown = \( 3000 m \) in \( 15 s = 200 m/s = 720 \) km/h.

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 21: From an aeroplane vertically above a straight horizontal road, the angles of depression of two stones 1 km apart on opposite sides of the aeroplane are alpha and beta. The height of the aeroplane above the road, in km, is
(a) \(\dfrac{\tan\alpha+\tan\beta}{\tan\alpha\tan\beta}\)
(b) \(\tan\alpha+\tan\beta\)
(c) \(\dfrac{\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}\)
(d) \(\dfrac{\tan\alpha}{\tan\beta}\)
Show Answer & Explanation

Answer: (c) \(\dfrac{\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}\)

Explanation:
1. Let the height be \( h \): the distances are \( h \cot \alpha \) and \( h \cot \beta \).
2. \( h(\cot \alpha + \cot \beta) = 1 \), so \( h = 1/(\cot \alpha + \cot \beta) = \tan \alpha \tan \beta/(\tan \alpha + \tan \beta) \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

Question 22: A vertical tower is surmounted by a flagstaff of height h. At a point on the plane the angles of elevation of the bottom and the top of the flagstaff are alpha and beta. The height of the tower is
(a) \(\dfrac{h\tan\alpha}{\tan\beta-\tan\alpha}\)
(b) \(\dfrac{h\tan\beta}{\tan\beta-\tan\alpha}\)
(c) \(\dfrac{h\tan\alpha}{\tan\beta+\tan\alpha}\)
(d) \(h(\tan\beta-\tan\alpha)\)
Show Answer & Explanation

Answer: (a) \(\dfrac{h\tan\alpha}{\tan\beta-\tan\alpha}\)

Explanation:
1. Let the tower be \( H \) and the distance \( d \): \( H = d \tan \alpha \) and \( H + h = d \tan \beta \).
2. Subtract: \( h = d(\tan \beta - \tan \alpha) \), so \( H = h \tan \alpha/(\tan \beta - \tan \alpha) \).

Teacher's Note:
1. Draw the right triangle first and mark the angle of elevation or depression from the horizontal line.
2. Use \( \tan \) for height over horizontal distance and keep the unknown in one equation.

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

Chapter MCQs with Answers for Class 10 Mathematics

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FAQs

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