Answer
$\angle{C} \approx 31.7^{\circ}$
Work Step by Step
Consider a triangle $\text{ABC}$.
Apply the Law of Sines.
$\dfrac{\sin A}{a} =\dfrac{\sin B}{b} =\dfrac{\sin C}{c} $
Rearrange the above two ratio setup to obtain:
$\dfrac{\sin A}{a} =\dfrac{\sin C}{c}\\\dfrac{\sin 52^{\circ}}{15} =\dfrac{\sin C}{10} \\ \sin C=\dfrac{10 \sin 52^{\circ}}{15} ~~~(1)$
In order to calculate $\angle{C}$, we need to rearrange the equation (1) as follows:
$\angle{C} =\sin^{-1} (\dfrac{10 \sin 52^{\circ}}{15}) \approx 31.7^{\circ}$
