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Problem 546

AMC 12 late, AIME early

Geometry Difficulty 4.9 Find the answer

3. As shown in Figure 1, in the pyramid $P-ABCD$ , $AB$ $/ / CD, \angle ADC=90^{\circ}$ , $PD \perp$ plane $ABCD$ . If $PD=AD=AB=1$ , $CD=2$ , then the sine value $\sin \theta$ of the plane angle $\theta$ formed by the dihedral angle between plane $PAD$ and plane $PBC$ is ( ).
(A) $\frac{\sqrt{7}}{3}$
(B) $\frac{\sqrt{30}}{6}$
(C) $\frac{\sqrt{31}}{6}$
(D) $\frac{2 \sqrt{2}}{3}$

Your answer

Official solution

3. B.

Given $A B / / C D$ and $\angle A D C=90^{\circ}$ , we get $\angle D A B=$ $90^{\circ}$ . Connect $B D$ . By the Pythagorean theorem, we have $B D=\sqrt{2}, P B=$ $\sqrt{3}, P C=\sqrt{5}, B C=\sqrt{2}$ (obviously, $\angle P D B=90^{\circ}$ ).

Notice that $B C^{2}+P B^{2}=P C^{2}$ , so $\angle P B C$ $=90^{\circ}$ . Also, the projection of $\triangle P B C$ on the plane $P A D$ is exactly $\triangle P A D$ , thus,
$\cos \theta=\frac{S_{\triangle P A D}}{S_{\triangle P B C}}=\frac{\frac{1}{2} \times 1^{2}}{\frac{1}{2} \times \sqrt{2} \times \sqrt{3}}=\frac{1}{\sqrt{6}} .$

Therefore, $\sin \theta=\sqrt{1-\cos ^{2} \theta}=\frac{\sqrt{30}}{6}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.