Question

In Δ ABC, if cos A = sin B - cos C then show that it is a right-angled triangle.

Answer 1

cos A = sin B - cos C

∴ cos A + cos C = sin B

∴ `2 cos (("A + C")/2). cos (("A - C")/2)` = sin B

∴ `2 cos (pi/2 - "B"/2). "cos" (("A - C")/2)` = sin B .....[∵ A + B + C = π]

∴ `2 sin  "B"/2. cos (("A - C")/2) = 2 sin  "B"/2. "cos" "B"/2`

∴ `cos  (("A - C")/2) = cos  "B"/2`

∴ `("A - C")/2 = "B"/2`

∴ A - C = B

∴ A = B + C

∴ A + B + C = 180° gives

∴ A + A = 180°

∴ 2A = 180°

∴ A = 90°

∴ Δ ABC is a right angled triangle.