A square pyramid V–ABCD has base side a>0. Face VAD is equilateral and perpendicular to the base; M is the midpoint of AD. Find the tangent of the dihedral angle between VAD and VDB.

01 / Standard#Your turn

Official paper · jm02-2025 · 1(c) · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Choose the midpoint P of the common edge VD.

Hint 2

Find lines AP and BP perpendicular to VD.

Worked solution

In equilateral VAD, AP⊥VD. Also VB=BD=√2a, so the median BP of isosceles VBD is perpendicular to VD.
Thus ∠APB measures the dihedral angle. AB is perpendicular to face VAD, so triangle APB is right at A.
$AP=\frac{\sqrt3}2a,\quad AB=a$
Use the opposite-to-adjacent ratio at P.
$\tan x=\frac{AB}{AP}=\frac2{\sqrt3}=\frac{2\sqrt3}3$

tan x=2√3/3.

Checks and common pitfalls: A dihedral angle requires two lines perpendicular to the same common edge.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
In equilateral VAD, AP⊥VD. Also VB=BD=√2a, so the median BP of isosceles VBD is perpendicular to VD.
Thus ∠APB measures the dihedral angle. AB is perpendicular to face VAD, so triangle APB is right at A.
$AP=\frac{\sqrt3}2a,\quad AB=a$
Use the opposite-to-adjacent ratio at P.
$\tan x=\frac{AB}{AP}=\frac2{\sqrt3}=\frac{2\sqrt3}3$

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

2025 JM02