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 1, point $K$ lies on the line containing the diagonal of the base $AC$, and $KC = KA + AC$. The lateral edge $TC$ is $2 \sqrt{2}$, 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 lateral edge $TA$ and the plane of the lateral face $TCD$.

(20 points)
# Solution:

$K N \perp A B, N=K N \cap A B$,

1) 

$K M \perp C D, M=K M \cap C D . \quad K N=\frac{T K}{\operatorname{tg} 60^{\circ}}=\frac{T K \sqrt{3}}{3}$,

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

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

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

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

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

$a=\frac{T K \cdot K C^{2}}{\sqrt{3} K M \cdot M C}=\frac{T C^{2}-T K^{2}}{3 \sqrt{T C^{2}-4 T K^{2}}}=\frac{7}{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f8bb21871417ffa2deg-4.jpg?height=905&width=754&top_left_y=1095&top_left_x=1188)

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

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

$A P \perp F L, \quad A P \perp T D C, T P$ - projection of $T A$ on plane $T D C$, angle $A T P$ is the desired angle. $\angle A T P=\varphi$, $\sin \varphi=\frac{A P}{T A}=\frac{\sqrt{3} T K}{3 \sqrt{T K^{2}+A K^{2}}}=\frac{\sqrt{3} T K}{\sqrt{8 T K^{2}+T C^{2}}}=\frac{\sqrt{3} T K}{\sqrt{8 T K^{2}+T C^{2}}}=\frac{\sqrt{3}}{4}, \varphi=\arcsin \frac{\sqrt{3}}{4}$.

Answer: $\frac{7}{6}, \quad \arcsin \frac{\sqrt{3}}{4}$.