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Question
In ΔABC, if a cos A = b cos B then prove that the triangle is either a right angled or an isosceles traingle.
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Solution
Using the sine rule,
`a/"sinA" = b/"sinB"` = k
a = k sin A and b = k sin B
∴ a cos A = b cos B gives
k sinA cosA = k sinB cosB
∴ 2sinA cosA = 2sinB cosB
∴ sin 2A = sin 2B
∴ sin2A – sin2B = 0
∴ 2cos(A + B).sin(A – B) = 0
∴ 2cos(π – C).sin(A – B) = 0 ...[∵ A + B + C = π]
∴ - 2cosC. sin(A – B) = 0
∴ cosC = 0 OR sin(A – B) = 0
∴ C = 90° OR A – B = 0
∴ C = 90° OR A = B
∴ the triangle is either rightangled or an isosceles triangle.
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