Maths Class 10 Chapter 9: Some Applications of Trigonometry

# Maths Class 10 Chapter 9: Some Applications of Trigonometry

Important Concepts and Formulas

1.      In height and distance, the following terms are used:

I.            The line of sight is the line drawn from the eye of an observer to the point in        the object viewed by the observer.
II.            The angle of elevation of an object viewed is the angle formed by the line of        sight with the horizontal line when it is above the horizontal level, i.e., the            case when we raise our head to look at the object.
III.            The angle of depression of an object viewed is the angle formed by the line of       sight with the horizontal line when it is below the horizontal level, i.e., the             case when we lower our head to look at the object.

2.      The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.

## Solved Examples on Applications of Trigonometry

Example 1: A tower AB stands vertically on the ground. From a point C on the ground, which is 12 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

Solution: From the given figure, it is clear that AB is the tower and ACB is the angle of elevation.

Thus, ACB = 60° and BC = 12 m.
To solve this problem, we use the trigonometric ratio tan 60° (or cot 60°), as the ratio involves height AB and base BC.
Now, tan 60° = AB/BC
i.e., √3 = AB/12
i.e., AB = 12√3 m
Hence, the height of the tower is 12√3 m.

Example 2: An observer 1.7 m tall is 35 m away from a chimney. The angle of elevation of the top of the chimney from her eye is 45°. What is height of the chimney.

Solution: In the given figure, AB is the chimney, CD the observer and ADE the angle of elevation.
CD = BE = 1.7 m

In this case, ADE is a triangle, right-angled at E and we are required to find the height of the chimney.
We have AB = AE + BE = AE + 1.7
And DE = BC = 35 m
To find AE, we choose a trigonometric ratio tan 45°, which involves the perpendicular AE and base DE.
Thus, tan 45° = AE/ DE
i.e., 1 = AE/35
Therefore, AE = 35 m
So, the height of the chimney AB = AE + BE = 35 + 1.7 = 36.7 m

Example 3: A flag is hoisted at the top of a building 20 m high. From a point P on the ground, the angle of elevation of the top of the building and the angle of elevation of the top of the flag are 30° and 45° respectively. Find the length of the flag and the distance of the building from the point P.

Solution: In the given figure, AB represents the height of the building and BD the height of the flag. We have two right-angled triangles PAB and PAD.

We are required to find the length BD and the distance PA.
Since we know the height of the building AB, we first consider the triangle PAB.
We have,              tan 30° = AB/AP
i.e.,   1/√3 = 20/AP
Therefore, AP = 20√3 m
Hence, the distance of the building from P is 20√3 m.
Suppose BD = x m and then AD = (20 + x) m.
We have,         tan 45° = AD/AP
i.e., 1 = (20 + x)/20√3
i.e., 20 + x = 20√3
i.e., x = 20√3 – 20 = 20(√3 – 1) m
Hence, the length of the flag is 20(√3 – 1) m.

Example 4: The shadow of a tower standing on the ground is found to be 50 m longer when the Sun’s altitude is 30° than when it is 45°. Find the height of the tower.

Solution: In the given figure, AB is the tower and BC is the length of the shadow when the Sun’s altitude is 45° and BD is the length of the shadow when the Sun’s altitude is 30°.

Now, let us suppose AB = h m and BC = x m
CD = 50 m (given)
Thus, BD = (50 + x) m
In Î”ABC, tan 45° = AB/BC
or, 1 = h/x
or, h = x     ---- (1)
In Î”ABD, tan 30° = AB/BD
or, 1/√3 = h/(50 + x)
or, h√3 = 50 + x    ---- (2)
From (1) and (2), we get
x√3 = 50 + x
x√3 – x = 50
x(√3 – 1) = 50
x(1.732 – 1) = 50
x(0.732) = 50
x = 50/0.732
x = 68.3 m (Approx.)

Hence, the height of the tower is approximately 68.3 m.