A,B,C,D are coplanar as in the official diagram, with B and D on opposite sides of AC. ABC is equilateral; PAD is right isosceles at A; ∠PAC=π/2, ∠PCA=π/6, PA=1 and CD=2. Find the cosine of the dihedral angle between APD and CPD.

01 / Standard#Your turn

Official paper · jm02-2024 · 1(c) · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Let M be the midpoint of their common edge PD.

Hint 2

AM and CM are both perpendicular to PD.

Worked solution

The equal sides in each triangle make its median an altitude.
$PA=DA=1,\quad PC=\sqrt{1+3}=2=DC,\quad AM\perp PD,\ CM\perp PD$
Compute lengths using right triangles.
$PD=\sqrt2,\quad AM=\frac{\sqrt2}2,\quad CM=\sqrt{4-1/2}=\frac{\sqrt{14}}2$
Use the cosine rule in AMC.
$\cos\angle AMC=\frac{AM^2+CM^2-AC^2}{2AM\,CM}=\frac{1/2+7/2-3}{\sqrt7}=\frac{\sqrt7}7$

Cosine √7/7.

Checks and common pitfalls: Use the plane angle perpendicular to the common edge, not an arbitrary face angle.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The equal sides in each triangle make its median an altitude.
$PA=DA=1,\quad PC=\sqrt{1+3}=2=DC,\quad AM\perp PD,\ CM\perp PD$
Compute lengths using right triangles.
$PD=\sqrt2,\quad AM=\frac{\sqrt2}2,\quad CM=\sqrt{4-1/2}=\frac{\sqrt{14}}2$
Use the cosine rule in AMC.
$\cos\angle AMC=\frac{AM^2+CM^2-AC^2}{2AM\,CM}=\frac{1/2+7/2-3}{\sqrt7}=\frac{\sqrt7}7$

Think first. Reveal a hint when the class is ready.

2024 JM02