Trigonometric Ratios
Suppose a boy is standing in front of a lamp post at a certain distance. The height of the boy is 170 cm and the length of his shadow is 150 cm.
You can see from the above figure that the boy and his shadow form a right-angled triangle as shown in the figure below.
The ratio of the height of the boy to his shadow is 170:150 i.e., 17:15.
Is this ratio related to either of the angles of ΔABC?
We can also conclude the following:
cos A , tan A
Also, note that
tan A = and cot A =
Let us now solve some more examples based on trigonometric ratios.
Example 1: In a triangle ABC, right-angled at B, side AB = 40 cm and BC = 9 cm. Find the value of sin A, cos A, and tan A.
Solution:
It is given that AB = 40 cm and BC = 9 cm
Using Pythagoras theorem in ΔABC, we obtain
(AC)2 = (AB)2 + (BC)2
(AC)2 = (40)2 + (9)2
(AC)2 = 1600 + 81
(AC)2 = 1681
(AC)2 = (41)2
AC = 41 cm
Now, sin A
cos A
tan A
Example 2: From the given figure, find the values of cosec C and cot C, if AC = BC + 1.
Solution:
Now, it is given that AB = 5 cm and
AC = BC + 1 … (1)
By Pythagoras theorem, we obtain
(AB)2 + (BC)2 = (AC)2
⇒ (AC)2 − (BC)2 = (AB)2
⇒ (BC + 1)2 – (BC)2 = (5)2 [Using (1)]
⇒ (BC)2 + 1 + 2BC – (BC)2 = 25
⇒ 2BC = 25 – 1
⇒ 2BC = 24
⇒ BC = 12 cm
∴ AC = 12 + 1 = 13 cm
Thus, cosec C =
=
And, cot C =
=
Example 3: In a right-angled triangle ABC, which is right-angled at B, tan A . Find the value of cos A and sec A.
Solution:
It is given that tan A
We know that tan A
⇒
Let BC = 12k and AB = 5k
Using Pythagoras theorem in ΔABC, we obtain
(AC)2 = (AB)2 + (BC)2
= (5k)2 + (12k)2
= 25k2 + 144k2
(AC)2 = 169k2
AC = 13k
Now, cos A
Sec A
Use Trigonometric Ratios In Solving Problems
If we know the value of a trigonometric ratio, then we can find the values of other trigonometric ratios and the value of any expression involving these trigonometric ratios.
Example1: If cot A , then find the value of .
Solution:
It is given that cot A .
We know that
tan A
tan A = 3
Then,
Example 2: Find the value of (sin2 θ + cos2 θ), if sec θ .
Solution:
We have sec θ … (i)
We know that sec θ
Let ABC be a triangle in which ∠C = θ, therefore we have
sec θ
From equations (i) and (ii), we have
Let us take AC = 25k and BC = 24k.
Using Pythagoras theorem, we have,
(AC)2 = (AB)2 + (BC)2
(25k)2 = (AB)2 + (24k)2
625k2 = (AB)2 + 576k2
AB2 = (625 − 576)k2
AB2 = 49k2
AB = 7k
∴ sin θ
cos θ
Now, sin2 θ + cos2 θ
= 1
∴ sin2 θ + cos2 θ = 1