Consider a right angled triangle $A B C$ , in which $∠ A B C = 90^{o}$ . Write all trigonometric ratios with respect to $∠ A$ and $∠ C$ .

Open in App

Solution

Here $∠ C A B$ is an acute angle. Observe the position of the sides with respect to angle $A$ .
- $B C$ is the opposite side of Angle $A$ .
- $A B$ is the adjacent side with respect to angle $A$ .
- $A C$ is the hypotenuse of the right angled triangle $A B C$ .
The trigonometric ratios of the angle $A$ in the right angled triangle $A B C$ can be defined as follows.
$sin A = \frac{S i d e o p p o s i t e t o a n g l e A}{H y p o t e n u s e} = \frac{B C}{A C}$
$cos A = \frac{S i d e a d j a c e n t t o a n g l e A}{H y p o t e n u s e} = \frac{A B}{A C}$
$tan A = \frac{S i d e o p p o s i t e t o a n g l e A}{S i d e a d j a c e n t t o a n g l e A} = \frac{B C}{A B}$
$csc A = \frac{H y p o t e n u s e}{S i d e o p p o s i t e t o a n g l e A} = \frac{A C}{B C}$
$sec A = \frac{H y p o t e n u s e}{S i d e a d j a c e n t t o a n g l e A} = \frac{A C}{A B}$
$cot A = \frac{S i d e a d j a c e n t t o a n g l e A}{S i d e o p p o s i t e t o a n g l e A} = \frac{A B}{B C}$
Now let us define the trigonometric ratios for the acute angle $C$ in the right angled triangle, $∠ A B C = 90^{o}$
Observe that the position of the sides changes when we consider angle ' $C$ ' in place of angle $A$ .
$sin C = \frac{A B}{A C}$
$cos C = \frac{B C}{A C}$
$tan C = \frac{A B}{B C}$
$csc C = \frac{A C}{A B}$
$sec C = \frac{A C}{B C}$
$cot C = \frac{B C}{A B}$

Suggest Corrections

Similar questions

A

Δ A B C

△ A B C

B, A B = 10

A C = 26

A

C

△ A B C

∠ B = 90^{o}, ∠ C = 50^{o}

sin 50^{o}

cos 50^{o}

Δ A B C, ∠ A = 90^{o}, a = 25 c m, b = 7 c m

∠ B

∠ C

In a right $△$ ABC, if $∠$ A is acute and tanA = 3/4, find the remaining trigonometric ratios of $∠$ A.

Join BYJU'S Learning Program