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