 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-4/11e/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

\frac{s i n h a}{s i n α} = \frac{s i n h b}{s i n β} = \frac{s i n h c}{s i n γ} .

γ = π /2

in the cosine laws for sides and angles and eliminate

cos α

and

cos β

. Straight algebra gives the right-triangle identities

tan α = \frac{sinh a}{cosh a sinh b}, tan α tan β cosh c = 1 .

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