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.

Question

Quizplus · Accepted Answer

Given the formula for the area of a triangle $A = \frac{1}{2}ab\sin C$ and the values $A = 15$, $a = 5$, and $b = 8$, we can solve for $\sin C$ and then find $C$. Plugging in the values, we get $15 = \frac{1}{2} \cdot 5 \cdot 8 \cdot \sin C$, which simplifies to $\sin C = \frac{15}{20}$ or $\sin C = 0.75$. Solving for $C$ gives $C \approx 48.59^\circ$, which is an acute angle.

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