Answer
$\sin\theta$ = $\frac{12}{13}$
Work Step by Step
We know from first Pythagorean identity that-
$\sin\theta$ = ± $\sqrt (1-\cos^{2}\theta)$
As $\theta$ terminates in Q I, Therefore $\sin\theta$ will be positive-
$\sin\theta$ = $\sqrt (1-\cos^{2}\theta)$
substitute the given value of $\cos\theta$-
$\sin\theta$ = $\sqrt (1-(\frac{5}{13})^{2})$
$\sin\theta$ = $\sqrt (1-\frac{25}{169})$
$\sin\theta$ = $\sqrt (\frac{169 - 25}{169})$ = $\sqrt (\frac{144}{169})$
$\sin\theta$ = $\frac{12}{13}$
