6. The base of the pyramid $T A B C D$ is a rhombus $A B C D$. The height of the pyramid $T K$ is 5, point $K$ lies on the line containing the diagonal of the base $A C$, and $K C = K A + A C$. The lateral edge $T C$ is $6 \sqrt{5}$, and the lateral faces are inclined to the plane of the base at angles of $30^{\circ}$ and $60^{\circ}$. Find the length of the side of the base and the angle between the side of the base $A B$ and the lateral face $T B C$.
# Solution.

1) $\begin{aligned} & K N \perp A D, N=K N \cap A D, \\ & K M \perp C B, M=K M \cap C B .\end{aligned} \quad K N=\frac{T K}{\tan 60^{\circ}}=\frac{T K \sqrt{3}}{3}$,

$K M=\frac{T K}{\tan 30^{\circ}}=T K \sqrt{3}, \quad N M=\frac{2 \sqrt{3}}{3} T K$.

$K C=\sqrt{T C^{2}-T K^{2}}$,

$M C=\sqrt{K C^{2}-K M^{2}}=\sqrt{T C^{2}-4 T K^{2}}$.

2) $\angle A C B=\alpha, \sin \alpha=\frac{K M}{K C}, \quad \cos \alpha=\frac{M C}{K C}$,

$\sin 2 \alpha=\frac{2 K M \cdot M C}{K C^{2}}, \quad a=A B=\frac{M N}{\sin 2 \alpha}$,

$a=\frac{T K \cdot K C^{2}}{\sqrt{3} K M \cdot M C}=\frac{T C^{2}-T K^{2}}{3 \sqrt{T C^{2}-4 T K^{2}}}=\frac{31 \sqrt{5}}{12}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f8bb21871417ffa2deg-8.jpg?height=825&width=685&top_left_y=370&top_left_x=1211)

2) $A F \| T K, \quad F \in T C, \quad K A: A C=1: 2, \quad F A=\frac{2 T K}{3}$.

$A L \perp M C, \quad L \in M C, \quad A L=M N=\frac{2 \sqrt{3}}{3} T K, \quad F L=\sqrt{F A^{2}+A L^{2}}=\frac{4}{3} T K$.

$A P \perp F L, \quad A P \perp T B C, B P$ - projection of $A B$ on the plane $T B C$, angle $A B P$ is the desired angle. $\angle A B P=\varphi$, $\sin \varphi=\frac{A P}{A B}=\frac{\sqrt{3} T K}{3 a}=\frac{4 \sqrt{15}}{31}, \varphi=\arcsin \frac{4 \sqrt{15}}{31}$.

Answer: $\frac{31 \sqrt{5}}{12}, \quad \arcsin \frac{4 \sqrt{15}}{31}$.