[ Dihedral angle ] [Theorem of three perpendiculars ]

## [ Volume of a tetrahedron and pyramid ]

The base of the pyramid $S A B C D$ is a rectangle $A B C D$, where $A B=a, A D=b$; $S C$ is the height of the pyramid, $C S=h$. Find the dihedral angle between the planes $A B S$ and $A D S$.
Let's first prove the following statement: if $V$ is the volume of a tetrahedron, $S_1$ and $S_2$ are the areas of two faces, $l$ is the length of their common edge, and $\phi$ is the dihedral angle between them, then $V=\frac{2}{3} \cdot \frac{S_{1} S_{2} \sin \phi}{4}$. Let the edge $MN$ of the tetrahedron $KLMN$ be $l$, the angle between the faces $MNL$ and $MNK$ be $\phi$, $S_{\triangle MNK}=S_1$, and $S_{\triangle MNL}=S_2$ (Fig.1). If $LH$ is the height of the tetrahedron dropped to the base $MNK$, and $HP$ is the perpendicular dropped from point $H$ to $MN$, then by the theorem of three perpendiculars, $LP \perp MN$, so $LPH$ is the linear angle of the dihedral angle of the tetrahedron at edge $MN$. Therefore, $\angle LPH=\phi$. Then,

$$
V=V_{KLMN}=\frac{1}{3} S_{\triangle MNK} \cdot LH=\frac{1}{3} S_1 \cdot LP \sin \phi=\frac{1}{3} S_1 \cdot \frac{2 S_2}{l} \cdot \sin \phi=\frac{2}{3} \cdot \frac{S_1 S_2 \sin \phi}{l}
$$

The statement is proved. Now let's move on to our problem (Fig.2). From the right triangles $SCD$, $SCB$, and $SCA$, we find that

$$
\begin{gathered}
SD=\sqrt{CD^2 + SC^2}=\sqrt{a^2 + h^2}, SB=\sqrt{BC^2 + SC^2}=\sqrt{b^2 + h^2} \\
AS=\sqrt{AC^2 + SC^2}=\sqrt{a^2 + b^2 + h^2}
\end{gathered}
$$

From the theorem of three perpendiculars, it follows that $AB \perp BS$ and $AD \perp DS$. Then,

$$
S_{\triangle ABS}=\frac{1}{2} AB \cdot BS=\frac{1}{2} a \sqrt{b^2 + h^2}, S_{\triangle ADS}=\frac{1}{2} AD \cdot DS=\frac{1}{2} b \sqrt{a^2 + h^2}
$$

Let $\phi$ be the desired angle, and apply the previously proven statement to the tetrahedron $SABD$. We get

$$
V_{SABD}=\frac{2}{3} \cdot \frac{S_{\triangle ABS} \cdot S_{\triangle ADS} \sin \phi}{AS}
$$

or

$$
\frac{1}{3} \cdot \frac{1}{2} ab \cdot h=\frac{2}{3} \cdot \frac{\frac{1}{2} a \sqrt{b^2 + h^2} \cdot \frac{1}{2} b \sqrt{a^2 + h^2} \sin \phi}{\sqrt{a^2 + b^2 + h^2}}
$$

From this, we find that

$$
\sin \phi=\frac{h \sqrt{a^2 + b^2 + h^2}}{\sqrt{a^2 + h^2} \sqrt{b^2 + h^2}}
$$

Note that the projection of point $B$ onto the plane $ADS$ lies outside the triangle $ADS$, so $\phi > 90^\circ$. Therefore,

## Answer

$180^\circ - \arcsin \frac{h \sqrt{a^2 + b^2 + h^2}}{\sqrt{a^2 + h^2} \sqrt{b^2 + h^2}}$