Question

In $\sin \theta =\frac{5}{13}$, where $\theta$ is an acute angle, find the value of $\cos \theta$

Answer 1

Given: $\sin \theta =\frac{5}{13}$
To find : $\cos \theta =?$
We know that,
$\sin \theta =\frac{P}{H}=\frac{AB}{AB}=\frac{5}{13}$
According to Pythagoras theorem,
$P^2+B^2=H^2$
$(5{)}^2+B^2=(13{)}^2$
$B^2=169-25$
$B^2=144$
$B=12=BC$
$\cos \theta =\frac{B}{H}=\frac{BC}{AC}=\frac{12}{13}$
$\cos \theta =\frac{12}{13}$