Question

Right triangle XYZ has right angle Z. If the sin(X)=1213 , what is the cos(X)

Answer 1

Given:

Right triangle XYZ has right angle Z.

$\sin(x)=\dfrac{12}{13}$

To find:

The value of $\cos x$.

Solution:

We know that,

$\sin^2(x)+\cos^2(x)=1$

$\cos^2(x)=1-\sin^2(x)$

$\cos(x)=\pm\sqrt{1-\sin^2x}$

For a triangle, all trigonometric ratios are positive. So,

$\cos(x)=\sqrt{1-\sin^2x}$

It is given that $\sin(x)=\dfrac{12}{13}$. After substituting this value in the above equation, we get

$\cos(x)=\sqrt{1-(\dfrac{12}{13})^2}$

$\cos(x)=\sqrt{1-\dfrac{144}{169}}$

$\cos(x)=\sqrt{\dfrac{169-144}{169}}$

$\cos(x)=\sqrt{\dfrac{25}{169}}$

On further simplification, we get

$\cos(x)=\dfrac{\sqrt{25}}{\sqrt{169}}$

$\cos(x)=\dfrac{5}{13}$

Therefore, the required value is $\cos(x)=\dfrac{5}{13}$.