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.
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.
Now,
consider the triangle PAD.
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 ΔABC, tan
45° = AB/BC
or, 1 = h/x
or, h = x ---- (1)
In ΔABD, 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.