12.386. The lateral face of a regular triangular pyramid $S A B C$ forms an angle $\alpha$ with the base plane. A plane is drawn through the side $B C$ of the base and a point $D$ on the lateral edge $A S$. Find the angle between this plane and the base plane, if $A D: D S=k$.
## Solution.

Let $SO$ be the height of the regular pyramid $SABC$ (Fig. 12.236), $K$ the midpoint of $BC$, $\triangle BCD$ a section of the pyramid, $\angle SKO = \alpha, AB = a$, and the required $\angle DKA = \gamma$.

Then $\angle SKD = \alpha - \gamma, AK = \frac{a \sqrt{3}}{2}, OK = \frac{a \sqrt{3}}{6}$, and from $\triangle SOK (\angle SOK = 90^\circ)$: $SK = \frac{OK}{\cos \angle SKO} = \frac{a \sqrt{3}}{6 \cos \alpha}$.
$BC \perp$ plane $ASK$, plane $BDC \perp$ plane $ASK$. In the plane $ASK$, draw perpendiculars $AL$ and $SM$ to the line $DK$. Then $AL \perp$ plane $BDC, SM \perp$ plane $BDC$. If $CM = h_1, AL = h_2$, then from $\triangle ALK (\angle ALK = 90^\circ)$: $h_2 = AK \sin \gamma = \frac{a \sqrt{3}}{2} \sin \gamma$.

In $\triangle SMK (\angle SMK = 90^\circ)$: $h_1 = SK \sin (\alpha - \gamma) = \frac{a \sqrt{3}}{6 \cos \alpha} \sin (\alpha - \gamma)$.

$\Delta SMD \sim \triangle AMD \Rightarrow$

$\frac{h_2}{h_1} = \frac{AD}{DS} \Leftrightarrow \frac{\frac{a \sqrt{3}}{2} \sin \gamma}{\frac{a \sqrt{3}}{6 \cos \alpha} \sin (\alpha - \gamma)} = k \Leftrightarrow \frac{3 \cos \alpha \sin \gamma}{\sin (\alpha - \gamma)} = k \Leftrightarrow$

$\Leftrightarrow 3 \cos \alpha \sin \gamma = k \sin \alpha \cos \gamma - k \cos \alpha \sin \gamma \Leftrightarrow (k + 3) \sin \gamma \cos \alpha = k \sin \alpha \cos \gamma \Leftrightarrow$

$\Leftrightarrow \tan \gamma = \frac{k}{k + 3} \tan \alpha; \gamma = \arctan\left(\frac{k}{k + 3} \tan \alpha\right)$

Answer: $\arctan\left(\frac{k}{k + 3} \tan \alpha\right)$.