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 ∠CDB using arccos.

01 / Standard#Your turn

Official paper · jm02-2022 · 1(b)(ii) · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Use DB²=DM²+MB².

Hint 2

Apply the cosine rule in CDB.

Worked solution

Find the three side lengths.
$DB=\sqrt{1+4}=\sqrt5,\quad CD=\sqrt2,\quad CB=3$
Evaluate the cosine opposite CB.
$\cos\angle CDB=\frac{2+5-9}{2\sqrt2\sqrt5}=-\frac1{\sqrt{10}}$

∠CDB=arccos(−1/√10).

Checks and common pitfalls: The negative cosine means the angle is obtuse.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Find the three side lengths.
$DB=\sqrt{1+4}=\sqrt5,\quad CD=\sqrt2,\quad CB=3$
Evaluate the cosine opposite CB.
$\cos\angle CDB=\frac{2+5-9}{2\sqrt2\sqrt5}=-\frac1{\sqrt{10}}$

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

2022 JM02