Question

The area of a triangle is 474. Two of the side lengths are 87 and 19 and the included angle is acute. Find the measure of the included angle, to the nearest tenth of a degree.

Let $x$ be the measure of the included angle. We can use the formula for the area of a triangle to find an equation in terms of $x$.
The area of a triangle is given by $\frac{1}{2}bh$, where $b$ is the length of the base and $h$ is the length of the altitude from the base to the opposite vertex. In this case, the base is 87 and the altitude is the length of the line segment from the vertex opposite the included angle to the base, which we can find using the Law of Sines:
$$\frac{h}{\sin x} = \frac{19}{\sin(180^\circ-x)}$$
$$h = \frac{19\sin x}{\sin(180^\circ-x)}$$
Substituting this expression for $h$ into the formula for the area of the triangle, we get
$$\frac{1}{2}bh = \frac{1}{2} \cdot 87 \cdot \frac{19\sin x}{\sin(180^\circ-x)} = 474$$
Solving this equation for $x$ is somewhat difficult, but we can use a calculator or computer to approximate the value of $x$. Using a calculator, we find that $x \approx 31.4^\circ$. Rounding to the nearest tenth of a degree, the measure of the included angle is $\boxed{31.4^\circ}$.