31. Given that $\alpha$ is an acute angle satisfying
$$
\sqrt{369-360 \cos \alpha}+\sqrt{544-480 \sin \alpha}-25=0
$$
find the value of $40 \tan \alpha$.
31. Answer: 30
Let $X=\sqrt{369-360 \cos \alpha}$ and $Y=\sqrt{544-480 \sin \alpha}$. Observe that
$$
\begin{array}{l}
X^{2}=12^{2}+15^{2}-2(12)(15) \cos \alpha \\
Y^{2}=12^{2}+20^{2}-2(12)(20) \cos \left(90^{\circ}-\alpha\right)
\end{array}
$$
and $15^{2}+20^{2}=25^{2}=(X+Y)^{2}$, so we can construct a right-angled triangle $A B D$ as shown.
In particular $\angle A B C=\alpha$ and $\angle C B D=90^{\circ}-\alpha$. We can check that $\triangle A C B$ is in fact similar to $\triangle A B D$. So $\angle A D B=\alpha$ and
$$
40 \tan \alpha=40 \times \frac{15}{20}=30
$$