In pyramid P–ABC, PC⊥plane ABC, PC=3, ∠BCA=π/2. D lies on AB and E on BC; CD=DE=√2, CE=2 and EB=1. Find the dihedral angle P–AB–C using arctan.

01 / Standard#Your turn

Official paper · jm02-2022 · 1(c) · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Let X be the foot from C onto AB.

Hint 2

PC⊥the base implies PX⊥AB, so ∠PXC is the plane angle.

Worked solution

Compute the base hypotenuse and altitude by equal area expressions.
$AB=\sqrt{(3/2)^2+3^2}=3\sqrt5/2,\quad CX=\frac{AC\,BC}{AB}=3/\sqrt5$
AB is perpendicular to PC and CX, hence to plane PCX and PX. In right triangle PCX, use the tangent ratio.
$\tan\angle PXC=\frac{PC}{CX}=\sqrt5\implies\angle PXC=\arctan\sqrt5$

Dihedral angle arctan√5.

Checks and common pitfalls: Use a plane angle formed by perpendiculars to the common edge AB.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Compute the base hypotenuse and altitude by equal area expressions.
$AB=\sqrt{(3/2)^2+3^2}=3\sqrt5/2,\quad CX=\frac{AC\,BC}{AB}=3/\sqrt5$
AB is perpendicular to PC and CX, hence to plane PCX and PX. In right triangle PCX, use the tangent ratio.
$\tan\angle PXC=\frac{PC}{CX}=\sqrt5\implies\angle PXC=\arctan\sqrt5$

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

2022 JM02