CBSE Class 10 Mathematics Some Applications of Trigonometry Notes Set 02

Applications of Trigonometry

One of the uses of trigonometry is to find the heights of objects and the distances between points without actually measuring them. Two terms are used in all such problems.

Line of sight. The line drawn from the eye of the observer to the object being viewed. Through the eye O draw a horizontal line OX. If the object P is above OX, \( \angle XOP \) is called the angle of elevation of P as seen from O. If P is below OX, \( \angle XOP \) is called the angle of depression of P as seen from O.

Clinometer. The instrument used to measure these angles is called a clinometer. It is made of a semicircular disc with a small tube fixed along the diameter and a plumb line fixed at the centre. The tube is pointed at the object and the angle is read from the scale where the plumb line crosses it. For an angle of elevation of the top of a tree, the reading is taken on the side of the sight line; for an angle of depression the tube is tilted downwards in the same way.

Solved Examples

Example 5. The angle of elevation of an aeroplane from a point on the ground is 45°. After a flight of 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane.
Answer:
Let E be the point on the ground, B the foot of the first position A (EB = y), and let the plane cover AC = BD = x in 15 s.
In \( \triangle AEB \): \( \tan 45^\circ = \dfrac{3000}{y} \Rightarrow y = 3000 \). In \( \triangle CED \): \( \tan 30^\circ = \dfrac{3000}{x + y} \Rightarrow x + y = 3000\sqrt{3} \).
\( x = 3000(\sqrt{3} - 1) = 3000 \times 0.732 = 2196 \) m.
Speed \( = \dfrac{2196}{15} = 146.4 \) m/s \( = 146.4 \times \dfrac{18}{5} = \) 527.04 km/h.

Check Your Progress

Question 1. Find the angle of elevation of the sun when the length of the shadow of a pole is \( \sqrt{3} \) times the height of the pole.
Answer:
\( \tan\theta = \dfrac{\text{height}}{\text{shadow}} = \dfrac{1}{\sqrt{3}} \Rightarrow \theta = \) 30°.

Question 2. The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 100 metres apart, find the height of the light house.
Answer:
Both ships are on the same side. With height h: \( h\cot 30^\circ - h\cot 45^\circ = 100 \Rightarrow h(\sqrt{3} - 1) = 100 \).
\( h = \dfrac{100}{\sqrt{3} - 1} = \) \( 50(\sqrt{3} + 1) \) m (about 136.6 m).

Question 3. Two pillars of equal height are on either side of a roadway which is 30 metres wide. At a point on the roadway between the pillars the elevations of the top of the pillars are 60° and 30°. Find the height of the pillars and the position of the point.
Answer:
Let the pillar height be h and the point be at distance a from the 60° pillar. \( a = \dfrac{h}{\sqrt{3}} \) and \( 30 - a = h\sqrt{3} \).
\( \dfrac{h}{\sqrt{3}} + h\sqrt{3} = 30 \Rightarrow \dfrac{4h}{\sqrt{3}} = 30 \Rightarrow h = \) \( \dfrac{15\sqrt{3}}{2} \) m. The point is \( a = \dfrac{15}{2} = 7.5 \) m from the 60° pillar (22.5 m from the other).

Question 4. A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°; when he retires 14 metres from the bank he finds the angle to be 30°. Find the height of the tree and the breadth of the river.
Answer:
Let the height be h and the breadth be b. \( b = \dfrac{h}{\sqrt{3}} \) and \( b + 14 = h\sqrt{3} \).
\( h\sqrt{3} - \dfrac{h}{\sqrt{3}} = 14 \Rightarrow \dfrac{2h}{\sqrt{3}} = 14 \Rightarrow h = \) \( 7\sqrt{3} \) m, and the breadth is 7 m.

Question 5. The upper part of a tree broken by the wind makes an angle of 60° with the ground and the distance from the foot to the point where the top of the tree meets the ground is 20 metres. What was the height of the tree?
Answer:
The broken part is the hypotenuse: \( \cos 60^\circ = \dfrac{20}{L} \Rightarrow L = 40 \) m. The standing part is \( 20\tan 60^\circ = 20\sqrt{3} \) m.
Height of the tree \( = 40 + 20\sqrt{3} = \) \( 20(2 + \sqrt{3}) \) m.