In the diagram, $A B C D E F G H$ is a rectangular prism. (Vertex $H$ is hidden in this view.) If $\angle A B F=40^{\circ}$ and $\angle A D F=20^{\circ}$, what is the measure of $\angle B F D$, to the nearest tenth of a degree?

![](https://cdn.mathpix.com/cropped/2024_04_17_c7ef3b511bfae79753a1g-2.jpg?height=431&width=586&top_left_y=842&top_left_x=1409)
Without loss of generality, suppose that $A F=1$.

We let $\alpha=\angle A B F=40^{\circ}$ and $\beta=\angle A D F=20^{\circ}$ and $\theta=\angle B F D$.

![](https://cdn.mathpix.com/cropped/2024_04_30_a838ccd738aa6ca69034g-04.jpg?height=429&width=585&top_left_y=1672&top_left_x=824)

Using the cosine law in $\triangle B F D$, we obtain:

$$
\begin{aligned}
B D^{2} & =F B^{2}+F D^{2}-2(F B)(F D) \cos \theta \\
\cos \theta & =\frac{F B^{2}+F D^{2}-B D^{2}}{2(F B)(F D)}
\end{aligned}
$$

Now, $\sin \alpha=\frac{A F}{F B}$. Since $A F=1$, then $F B=\frac{1}{\sin \alpha}$.

Similarly, $F D=\frac{1}{\sin \beta}$.

Also, $\tan \alpha=\frac{A F}{A B}$. Since $A F=1$, then $A B=\frac{1}{\tan \alpha}=\frac{\cos \alpha}{\sin \alpha}$.

Similarly, $A D=\frac{1}{\tan \beta}=\frac{\cos \beta}{\sin \beta}$.

By the Pythagorean Theorem, $B D^{2}=A B^{2}+A D^{2}=\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}+\frac{\cos ^{2} \beta}{\sin ^{2} \beta}$.

Therefore,

$$
\begin{aligned}
\cos \theta & =\frac{F B^{2}+F D^{2}-B D^{2}}{2(F B)(F D)} \\
& =\frac{\frac{1}{\sin ^{2} \alpha}+\frac{1}{\sin ^{2} \beta}-\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}-\frac{\cos ^{2} \beta}{\sin ^{2} \beta}}{2 \cdot \frac{1}{\sin \alpha} \cdot \frac{1}{\sin \beta}} \\
& =\frac{\sin ^{2} \beta+\sin ^{2} \alpha-\cos ^{2} \alpha \sin ^{2} \beta-\cos ^{2} \beta \sin ^{2} \alpha}{2 \sin \alpha \sin \beta} \\
& =\frac{\sin ^{2} \beta\left(1-\cos ^{2} \alpha\right)+\sin ^{2} \alpha\left(1-\cos ^{2} \beta\right)}{2 \sin \alpha \sin \beta} \\
& =\frac{\sin ^{2} \beta \sin ^{2} \alpha+\sin 2 \sin ^{2} \beta}{2 \sin \alpha \sin \beta} \\
& =\sin \alpha \sin \beta
\end{aligned}
$$

Since $\alpha=40^{\circ}$ and $\beta=20^{\circ}$, then $\cos \theta=\sin \left(40^{\circ}\right) \sin \left(20^{\circ}\right) \approx 0.21985$.

Therefore, $\theta=\cos ^{-1}\left(\sin \left(40^{\circ}\right) \sin \left(20^{\circ}\right)\right) \approx 77.3^{\circ}$.

## Team Problems