CBSE Class 10 Mathematics Some Applications of Trigonometry Notes Set 01

Introduction

Trigonometry is used in the life around us. It has wide range of applications like it is used in astronomy to calculate distance from the earth to the planets and stars, and it can also be used in geography and in navigation.

Trigonometry can also be used to solve crimes by a forensic scientist. Suppose a body is discovered at the bottom of a block of flats, then the forensic scientist can have an idea as to whether the deceased fell, jumped or was pushed by using the measured distance from the building with the help of some elementary geometry.

Trigonometry can also be used to construct maps, to determine the position of an island in relation to the longitudes and latitudes. In this chapter we will learn how to apply trigonometry to find the heights of various objects and distances between points, without actually measuring them.

Angle of Elevation and Angle of Depression

Line of sight: Line of sight is the line drawn from the eye of the observer to the point on the object viewed by the observer.

Angle of Elevation: When a point on an object is viewed by an observer and the point is above the horizontal level (in this case the observer has to raise his/her head to look at the object), then the angle formed by the line of sight with the horizontal level is known as the Angle of Elevation.

Angle of Depression: When a point on an object is viewed by an observer and the point is below the horizontal level (in this case the observer has to lower his/her head to look at the object), then the angle formed by the line of sight with the horizontal level is known as the Angle of Depression.

The observer is at A and the object is at B. The angle of depression of B from A and the angle of elevation of A from B are equal, because the two horizontal lines are parallel (alternate angles).

Knowledge Hub

1. The angle of elevation is equal to the angle of depression (alternate angles between the parallel horizontal lines).
2. The angle of elevation of the sun is known as the sun's altitude.
3. Trigonometric ratios can be used to find the height or length of an object, or the distance between two distant objects.

Solved Examples

Example 1. A ladder 15 m long leans against a wall making an angle of \( 60^\circ \) with the wall. Find the height of the wall from the point the ladder touches the wall.

Answer:
Let the height of the wall be h metres. In right \( \triangle ABC \), \( \frac{AC}{AB} = \cos 60^\circ \Rightarrow \frac{h}{15} = \cos 60^\circ \).
\( h = 15 \times \cos 60^\circ = 15 \times \frac{1}{2} = 7.5 \)
Hence the ladder touches the wall at a height of 7.5 m.

Example 2. A girl standing on a horizontal plane finds a kite flying at a distance of 100 m from her at an elevation of \( 30^\circ \). A boy standing on the roof of a 20 m high building finds the angle of elevation of the same kite to be \( 45^\circ \). Both the girl and the boy are on opposite sides of the kite. Find the distance of the kite from the boy.

Answer:
Let M be the kite, O the girl and P the boy; \( \angle AOM = 30^\circ \) and \( \angle MPN = 45^\circ \), where MA is perpendicular to the ground and N is the point of MA at the height of the roof.
In \( \triangle AOM \): \( \sin 30^\circ = \frac{AM}{100} \Rightarrow \frac{1}{2} = \frac{AM}{100} \Rightarrow AM = 50 \) m.
\( MN = AM - AN = 50 - 20 = 30 \) m.
In \( \triangle PMN \): \( \sin 45^\circ = \frac{MN}{MP} \Rightarrow \frac{1}{\sqrt{2}} = \frac{30}{MP} \Rightarrow MP = 30\sqrt{2} \) m.
Hence the distance of the kite from the boy is \( 30\sqrt{2} \) m.

Example 3. A boy from the top of a 100 m high tower sees a car moving towards the tower at an angle of depression of \( 30^\circ \). After some time, the angle of depression becomes \( 60^\circ \). Find the distance travelled by the car during this time.

Answer:
Let PQ be the tower of height 100 m, and let R and C be the first and second positions of the car. By alternate angles, \( \angle PRQ = 30^\circ \) and \( \angle PCQ = 60^\circ \). Let \( CR = x \) and \( QC = y \).
In \( \triangle PQC \): \( \frac{100}{y} = \tan 60^\circ = \sqrt{3} \Rightarrow y = \frac{100}{\sqrt{3}} \).
In \( \triangle PQR \): \( \frac{100}{x + y} = \tan 30^\circ = \frac{1}{\sqrt{3}} \Rightarrow x + y = 100\sqrt{3} \).
\( x = 100\sqrt{3} - \frac{100}{\sqrt{3}} = \frac{300 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} = \frac{200\sqrt{3}}{3} \).
Hence the car travels \( \frac{200\sqrt{3}}{3} \) m (about 115.5 m).

Example 4. A man standing on the deck of a ship, which is 10 m above the water level, observes the angle of elevation of the top of a cliff as \( 60^\circ \) and the angle of depression of the base of the cliff as \( 30^\circ \). Calculate the distance of the cliff from the ship and the height of the cliff. [CBSE 2016]

Answer:
Let M be the man on the deck, A the top and D the base of the cliff, and B the point of AD at the level of M. Then \( MB \parallel \) the water level, so \( \angle NDM = \angle BMD = 30^\circ \).
In right \( \triangle MND \): \( \tan 30^\circ = \frac{10}{ND} \Rightarrow ND = 10\sqrt{3} \approx 17.3 \) m. So \( BM = ND = 10\sqrt{3} \) m.
In right \( \triangle ABM \): \( \tan 60^\circ = \frac{AB}{BM} \Rightarrow \sqrt{3} = \frac{AB}{10\sqrt{3}} \Rightarrow AB = 30 \) m.
\( AD = AB + BD = 30 + 10 = 40 \) m.
The distance of the cliff from the ship is 17.3 m (\( 10\sqrt{3} \) m) and the height of the cliff is 40 m.

Example 5. From the top of a cliff, the angles of depression of two consecutive kilometre stones due east are found to be \( 45^\circ \) and \( 30^\circ \) respectively. Find the height of the cliff.

Answer:
Let the height of the cliff be h metres, and let P and Q be the two stones, \( PQ = 1000 \) m, with \( AP = x \) m, where AB is the cliff.
In \( \triangle ABP \): \( \tan 45^\circ = \frac{h}{x} \Rightarrow x = h \) ...(i)
In \( \triangle ABQ \): \( \tan 30^\circ = \frac{h}{x + 1000} \Rightarrow x + 1000 = \sqrt{3}h \).
Using (i): \( h + 1000 = \sqrt{3}h \Rightarrow h(\sqrt{3} - 1) = 1000 \Rightarrow h = \frac{1000}{\sqrt{3} - 1} = 500(\sqrt{3} + 1) \).
Hence the height of the cliff is \( 500(\sqrt{3} + 1) \) m (about 1366 m).

Example 6. The height of a building is h units and the angle of elevation of the top of the building is \( \alpha \). On moving a distance \( \frac{h}{2} \) units towards the building, the angle of elevation becomes \( \beta \). What is the value of \( (\cot \alpha - \cot \beta) \)?

Answer:
Let AM be the building of height h, with \( \angle ABM = \alpha \), \( \angle ACM = \beta \) and \( BC = \frac{h}{2} \).
In \( \triangle ACM \): \( \tan \beta = \frac{h}{CM} \Rightarrow CM = h\cot \beta \) ...(i)
In \( \triangle ABM \): \( \tan \alpha = \frac{h}{\frac{h}{2} + CM} \Rightarrow \frac{h}{2} + CM = h\cot \alpha \) ...(ii)
Putting (i) in (ii): \( \frac{h}{2} + h\cot \beta = h\cot \alpha \Rightarrow h(\cot \alpha - \cot \beta) = \frac{h}{2} \).
Hence \( \cot \alpha - \cot \beta = \) \( \frac{1}{2} \).

Example 7. The angle of elevation of the top of an unfinished pillar at a point 150 m from its base is \( 30^\circ \). If the angle of elevation at the same point is to be \( 45^\circ \), then the pillar has to be raised to a height of how many metres?

Answer:
Let B be the point, C the foot of the pillar, M the present top and A the top when finished; \( CB = 150 \) m.
In \( \triangle MCB \): \( \tan 30^\circ = \frac{MC}{150} \Rightarrow MC = 50\sqrt{3} \) m.
In \( \triangle ACB \): \( \tan 45^\circ = \frac{AC}{150} \Rightarrow AC = 150 \) m.
Height to be raised \( = AM = AC - MC = 150 - 50\sqrt{3} = 50(3 - \sqrt{3}) \) m (about 63.4 m).

Example 8. Two poles are x metres apart and the height of one is double that of the other. If, from the mid-point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary, then find the height (in metres) of the shorter pole.

Answer:
Let the shorter pole be CD of height h, the taller pole PQ of height 2h, and let O be the mid-point of PC, so \( PO = OC = \frac{x}{2} \). Let \( \angle DOC = \theta \); then \( \angle QOP = 90^\circ - \theta \).
In \( \triangle OCD \): \( \tan \theta = \frac{h}{x/2} = \frac{2h}{x} \) ...(i)
In \( \triangle OPQ \): \( \tan(90^\circ - \theta) = \cot \theta = \frac{2h}{x/2} = \frac{4h}{x} \) ...(ii)
Multiplying (i) and (ii), and using \( \tan \theta \cdot \cot \theta = 1 \): \( \frac{8h^2}{x^2} = 1 \Rightarrow h^2 = \frac{x^2}{8} \Rightarrow h = \frac{x}{2\sqrt{2}} \).
Hence the height of the shorter pole is \( \frac{x}{2\sqrt{2}} \) m.

Example 9. When the Sun is at \( 30^\circ \), the shadow of a tower standing on level ground is 40 m longer than when it is at \( 60^\circ \) altitude. Determine the tower's height.

Answer:
Let AB be the tower of height h m. When the sun's altitude is \( 60^\circ \) the shadow is \( BD = x \), and when it is \( 30^\circ \) the shadow is \( BC = x + 40 \).
In \( \triangle ABD \): \( \tan 60^\circ = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}} \) ...(i)
In \( \triangle ABC \): \( \tan 30^\circ = \frac{h}{x + 40} \Rightarrow x + 40 = \sqrt{3}h \).
Using (i): \( \frac{h}{\sqrt{3}} + 40 = \sqrt{3}h \Rightarrow h + 40\sqrt{3} = 3h \Rightarrow h = 20\sqrt{3} \).
Hence the height of the tower is \( 20\sqrt{3} \) m (about 34.6 m).