Solve the problem.
-The area of a triangle is given by $$\mathrm { A } = \frac { 1 } { 2 } \mathrm { ab } \sin \mathrm { C } \text {, }$$
where $a$ and $b$ are the lengths of two of the sides and $C$ is the included angle. If $A = 29$ in. $2 , a = 8$ in., $b = 9$ in., and $C$ is an acute angle, what must $C$ be? Give your answer in degrees to the nearest hundredth.
A) $26.83 ^ { \circ }$
B) $53.66 ^ { \circ }$
C) $126.34 ^ { \circ }$
D) No such triangle exists.

Question

Quizplus · Accepted Answer

To find the angle $C$, we use the given formula for the area of a triangle, $A = \frac{1}{2}ab\sin C$, and solve for $C$. Substituting the given values, we get $29 = \frac{1}{2} \cdot 8 \cdot 9 \cdot \sin C$. Simplifying, we find $\sin C = \frac{29}{36}$. Solving for $C$ gives $C \approx 53.66^\circ$, which is an acute angle.

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