6. [45] Let $A B C D$ be an isosceles trapezoid with $A B=1, B C=D A=5, C D=7$. Let $P$ be the intersection of diagonals $A C$ and $B D$, and let $Q$ be the foot of the altitude from $D$ to $B C$. Let $P Q$ intersect $A B$ at
R. Compute $\sin \angle R P D$.
Answer: $\frac{4}{5}$
Let $M$ be the foot of the altitude from $B$ to $C D$. Then $2 C M+A B=C D \Longrightarrow C M=3$. Then $D M=4$ and by the Pythagorean theorem, $B M=4$. Thus $B M D$ is a right isosceles triangle i.e. $\angle B D M=\angle P D C=\frac{\pi}{4}$. Similarly, $\angle P C D=\frac{\pi}{4}$. Thus $\angle D P C=\frac{\pi}{2}$, which means quadrilateral $P Q D C$ is cyclic. Now, $\sin \angle R P D=\sin \angle D C Q=\sin \angle M C B=\frac{4}{5}$.
An alternate solution is also possible:
Note that $A C \perp B D$ since $A B^{2}+C D^{2}=1^{2}+7^{2}=5^{2}+5^{2}=B C^{2}+D A^{2}$. Thus $P$ is the foot of the altitude from $D$ to $A C$. Since $D$ is on the circumcircle of $\triangle A B C$, line $P Q R$ is the Simson line of $D$. Thus $R$ is the foot from $D$ to $A B$. Then from quadrilateral $R A P D$ being cyclic we have $\angle R P D=\angle R A D$. So $\sin \angle R P D=\frac{4}{5}$.