Question

In ∆ABC, `sqrt(2)`AC = BC, sin A = 1, sin²A + sin²B + sin²C = 2, then ∠A = ?, ∠B = ?, ∠C = ?

Answer 1

sin A = 1   ...[Given]

But, sin 90° = 1

∴ sin A = sin 90°

∴ A = 90°

`sqrt(2)`AC = BC   ...[Given]

∴ `(AC)/(BC) = 1/sqrt(2)`   ...(i)

∴ `sin B = (AC)/(BC)`   ...(ii) [By definition]

∴ `sin B = 1/sqrt(2)`   ...[From (i) and (ii)]

But, `sin 45^circ = 1/sqrt(2)`

∴ sin B = sin 45°

∴ B = 45°

sin²A + sin²B + sin²C = 2   ...[Given]

∴ `(1)^2 + (1/sqrt(2))^2 + sin^2C = 2`

∴ `1 + 1/2 + sin^2C = 2`

∴ `sin^2C = 2 - 3/2`

∴ `sin^2C = 1/2`

∴ `sin C = 1/sqrt(2)`

But, `sin 45^circ = 1/sqrt(2)`

∴ sin C = sin 45°

∴ C = 45°

∴ ∠A = 90°, ∠B = 45°, ∠C = 45°