Answer
$\cos \theta=-\frac{3}{5}$
Work Step by Step
Given: $\sin \theta=\frac{4}{5}$
We know the Identity : $\sin ^{2} \theta+\cos ^{2} \theta=1$
By Substituting we get
$\left(\frac{4}{5}\right)^{2}+\cos ^{2} \theta=1$
$\cos ^{2} \theta=1-\left(\frac{4}{5}\right)^{2}$
$\cos ^{2} \theta=1-\frac{16}{25}$
$\cos ^{2} \theta=\frac{25-16}{25}$
$\cos ^{2} \theta=\frac{9}{25}$
$\cos \theta=\pm\sqrt{\frac{9}{25}}$
$\cos \theta=\pm \frac{3}{5}$
Given theta in quadrant $II$, then cosine must be negative.
Therefore $\cos \theta=-\frac{3}{5}$.
