Solve the Problem $$\mathrm { A } = \frac { 1 } { 2 } \mathrm { ab } \sin \mathrm { C } \text {, }$$

Solve the problem.
-The area of a triangle is given by
$$\mathrm { A } = \frac { 1 } { 2 } \mathrm { ab } \sin \mathrm { C } \text {, }$$
where $\mathrm { a }$ and $\mathrm { b }$ are the lengths of two of the sides and $\mathrm { C }$ is the included angle. If $\mathrm { A } = 15$ in. 2 , $\mathrm { a } = 5$ in., $\mathrm { b } = 8$ in., and $\mathrm { C }$ is an acute angle, what must $\mathrm { C }$ be? Give your answer in degrees to the nearest hundredth.

A) $24.30 ^ { \circ }$
B) $131.41 ^ { \circ }$
C) $48.59 ^ { \circ }$
D) No such triangle exists.

Correct Answer:

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