CBSE Class 10 Mathematics Applications of Trigonometry Assignment Set 13

Previous Year Problems (CBSE Board)

Question 1. From a point P on the ground, the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (Take \( \sqrt{3} = 1.73 \)) [2 marks, CBSE 2012]
Answer:
Distance \( PB = 10\cot 30^\circ = 10\sqrt{3} = 17.3 \) m.
Total height to the flag top \( = PB\tan 45^\circ = 10\sqrt{3} \). Flagstaff \( = 10\sqrt{3} - 10 = \) \( 10(\sqrt{3} - 1) = 7.3 \) m; distance of the building from P is \( 10\sqrt{3} = 17.3 \) m.
In simple words: Subtract the building height from the total height.

Teacher's Note:
1. tan 45° = 1, so the total height equals the distance.

Question 2. The horizontal distance between two poles is 15 m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30°. If the height of the second pole is 24 m, find the height of the first pole. [use \( \sqrt{3} = 1.732 \)] [3 marks, CBSE 2012]
Answer:
The drop from the top of the second pole to the top of the first is \( 15\tan 30^\circ = 5\sqrt{3} = 8.66 \) m.
Height of the first pole \( = 24 - 5\sqrt{3} = \) 15.34 m.
In simple words: Subtract the drop from 24 m.

Teacher's Note:
1. The angle of depression equals the angle of elevation from the lower top.

Question 3. The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30°. The distance of the car from the base of the tower (in m) is [1 mark, CBSE 2013]
(A) \( 25\sqrt{3} \)   (B) \( 50\sqrt{3} \)   (C) \( 75\sqrt{3} \)   (D) 150
Answer:
Ans. (C) \( 75\sqrt{3} \)
\( \tan 30^\circ = \dfrac{75}{d} \Rightarrow d = 75\sqrt{3} \) m.
In simple words: Depression at the top equals elevation at the car.

Teacher's Note:
1. Distance = height × cot 30°.

Question 4. Two ships are approaching a light-house from opposite directions. The angles of depression of the two ships from the top of the light-house are 30° and 45°. If the distance between the two ships is 100 m, find the height of the light-house. [Use \( \sqrt{3} = 1.732 \)] [3 marks, CBSE 2014]
Answer:
\( h\cot 30^\circ + h\cot 45^\circ = 100 \Rightarrow h(\sqrt{3} + 1) = 100 \).
\( h = \dfrac{100}{2.732} = \) 36.6 m.
In simple words: The ships are on opposite sides, so the distances add.

Teacher's Note:
1. Add the two distances.

Question 5. The angle of elevation of the top of a chimney from the foot of a tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40 m, find the height of the chimney. According to pollution control norms, the minimum height of a smoke emitting chimney should be 100 m. State if the height of the above mentioned chimney meets the pollution norms. What value is discussed in this question? [4 marks, CBSE 2014]
Answer:
Distance between the foot of the tower and the chimney: \( d = 40\cot 30^\circ = 40\sqrt{3} \) m.
Height of the chimney \( = d\tan 60^\circ = 40\sqrt{3} \times \sqrt{3} = \) 120 m.
Since 120 m is more than 100 m, the chimney meets the pollution norms. The value discussed is concern for the environment, that is, controlling air pollution.
In simple words: First find the distance from the tower.

Teacher's Note:
1. Depression of the foot equals the elevation from the foot of the tower.

Question 6. If the height of a vertical pole is \( \sqrt{3} \) times the length of its shadow on the ground, then the angle of elevation of the Sun at that time is [1 mark, CBSE 2014]
(A) 30°   (B) 60°   (C) 45°   (D) 75°
Answer:
Ans. (B) 60°
\( \tan\theta = \sqrt{3} \Rightarrow \theta = 60^\circ \).
In simple words: tan 60° = √3.

Teacher's Note:
1. Height over shadow is √3.

Question 7. A tower AB is 20 m high and BC, its shadow on the ground, is \( 20\sqrt{3} \) m long. Find the Sun's altitude. [1 mark, CBSE 2014]
Answer:
\( \tan\theta = \dfrac{20}{20\sqrt{3}} = \dfrac{1}{\sqrt{3}} \Rightarrow \theta = \) 30°.
In simple words: Height over shadow.

Teacher's Note:
1. tan 30° = 1/√3.

Question 8. The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the aeroplane is flying at a constant height of \( 1500\sqrt{3} \) m, find the speed of the plane in km/hr. [3 marks, CBSE 2015]
Answer:
Horizontal distances: \( 1500\sqrt{3}\cot 60^\circ = 1500 \) m and \( 1500\sqrt{3}\cot 30^\circ = 4500 \) m. The plane covers 3000 m in 15 s.
Speed \( = 200 \) m/s \( = 200 \times \dfrac{18}{5} = \) 720 km/h.
In simple words: Distance covered over time, then change to km/h.

Teacher's Note:
1. Multiply m/s by 18/5 for km/h.

Question 9. A ladder, leaning against a wall, makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder. [1 mark, CBSE 2015]
Answer:
\( \cos 60^\circ = \dfrac{2.5}{L} \Rightarrow L = \) 5 m.
In simple words: The foot distance is next to the angle.

Teacher's Note:
1. Use cosine.

Question 10. The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of the tower is 45°. Find the height of the tower PQ and the distance PX. (Use \( \sqrt{3} = 1.73 \)) [4 marks, CBSE 2016]
Answer:
Let PX = d and PQ = H. \( H = d\sqrt{3} \) and \( H - 40 = d \).
\( d(\sqrt{3} - 1) = 40 \Rightarrow d = 20(\sqrt{3} + 1) = \) 54.6 m, and \( H = \sqrt{3}d = 20\sqrt{3}(\sqrt{3} + 1) = \) 94.6 m.
In simple words: Both X and Y are the same distance from the tower.

Teacher's Note:
1. Y is 40 m above X, so the tower is 40 m taller than d.

Question 11. A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Find the distance of the hill from the ship and the height of the hill. [3 marks, CBSE 2016]
Answer:
Distance \( x = 10\cot 30^\circ = \) \( 10\sqrt{3} \) m.
Height of the hill above the man's eye \( = x\tan 60^\circ = 30 \) m. Height of the hill \( = 30 + 10 = \) 40 m.
In simple words: The hill base is 10 m below the eye.

Teacher's Note:
1. The hill height is the part above eye level plus 10 m.

Question 12. If a tower 30 m high casts a shadow \( 10\sqrt{3} \) m long on the ground, then what is the angle of elevation of the sun? [1 mark, CBSE 2017]
Answer:
\( \tan\theta = \dfrac{30}{10\sqrt{3}} = \sqrt{3} \Rightarrow \theta = \) 60°.
In simple words: Height over shadow.

Teacher's Note:
1. tan 60° = √3.

Question 13. On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower. [3 marks, CBSE 2017]
Answer:
\( \tan\theta = \dfrac{h}{4} \) and \( \tan(90^\circ - \theta) = \dfrac{h}{16} \). Multiplying: \( 1 = \dfrac{h^2}{64} \Rightarrow h = \) 8 m.
In simple words: Multiply the tangent and cotangent.

Teacher's Note:
1. tan θ × tan(90° − θ) = 1.

Question 14. An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively. Find the width of the river. [Use \( \sqrt{3} = 1.732 \)] [4 marks, CBSE 2017]
Answer:
The horizontal distances are \( 300\cot 45^\circ = 300 \) m and \( 300\cot 60^\circ = 100\sqrt{3} \) m on opposite sides.
Width \( = 300 + 100\sqrt{3} = 100\sqrt{3}(\sqrt{3} + 1) = \) 473.2 m.
In simple words: The two banks are on opposite sides, so the distances add.

Teacher's Note:
1. Use the exact value and then 1.732.