Under the same angle conditions, AC=5. Find the triangle area.

01 / Standard#Your turn

A+B+C=\pi,\quad\sin(C-A)=1,\quad\cos B=2\sqrt2/3,\quad AC=5

Official paper · jm01-2021 · II.5(b) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

C is obtuse because C=A+π/2.

Hint 2

Use the sine rule to find BC, then the two-side area formula.

Worked solution

Choose the correct cosine sign for the obtuse angle.
$\sin C=\sqrt{2/3},\quad\cos C=-1/\sqrt3,\quad\sin A=\sin(C-\pi/2)=1/\sqrt3$
Match opposite sides in the sine rule.
$BC=AC\frac{\sin A}{\sin B}=5\frac{1/\sqrt3}{1/3}=5\sqrt3$
The included angle between AC and BC is C.
$S=\frac12(5)(5\sqrt3)\sqrt{2/3}=\frac{25\sqrt2}2$

Area =25√2/2.

Checks and common pitfalls: Taking cos C positive would contradict the obtuse-angle condition.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Choose the correct cosine sign for the obtuse angle.
$\sin C=\sqrt{2/3},\quad\cos C=-1/\sqrt3,\quad\sin A=\sin(C-\pi/2)=1/\sqrt3$
Match opposite sides in the sine rule.
$BC=AC\frac{\sin A}{\sin B}=5\frac{1/\sqrt3}{1/3}=5\sqrt3$
The included angle between AC and BC is C.
$S=\frac12(5)(5\sqrt3)\sqrt{2/3}=\frac{25\sqrt2}2$

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

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