Question

Hi, can you help to solve this exercise please !!!

Answer 1

Given:-

$\sin \theta=\frac{12}{13}$

To find:-

The value of cos theta.

We know the formula,

$\sin ^2\theta+\cos ^2\theta=1$

From this formula, we derive the value of cos theta. we get,

$\begin{gathered} \sin ^2\theta+\cos ^2\theta=1 \\ \cos \theta=\sqrt[]{1-\sin ^2\theta} \end{gathered}$

Now we substitute the value of sin theta. so we get,

$\begin{gathered} \cos \theta=\sqrt[]{1-\frac{12^2}{13^2}} \\ \cos \theta=\sqrt[]{1-\frac{144}{169}} \\ \cos \theta=\sqrt[]{\frac{169-144}{169}} \\ \cos \theta=\sqrt[]{\frac{25}{169}} \\ \cos \theta=\frac{5}{13} \end{gathered}$

So the value of cos theta is,

$\frac{5}{13}$