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° = `1/sqrt(2)`

∴ sin B = sin 45°

∴ B = 45°

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

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

∴ `1 + 1/2 + sin^2"C"` = 2

∴ sin²C = `2 - 3/2`

∴ sin²C = `1/2`

∴ sin C = `1/sqrt(2)`

But, sin 45° = `1/sqrt(2)`

∴ sin C = sin 45°

∴ C = 45°

∴ ∠A = 90°, ∠B = 45°, ∠C = 45°