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Question
In ΔABC, if a cos A = b cos B, then prove that ΔABC is either a right angled or an isosceles triangle.
Solution
In ∆ABC by sine rule, we have
`"a"/(sin "A") = "b"/(sin "B")` = k
∴ a = k sin A and b = k sin B
Now, a cos A = b cos B ...[Given]
∴ k sin A cos A = k sin B cos B
∴ sin A cos A = sin B cos B
∴ 2 sin A cos A = 2 sin B cos B
∴ sin 2A = sin 2B
∴ sin 2A − sin 2B = 0
∴ 2 cos (A + B) sin (A − B) = 0
∴ 2 cos (π − C) sin (A − B) = 0 ...[∵ A + B + C = π]
∴ −2 cos C sin (A − B) = 0
∴ cos C = 0 or sin(A − B) = 0
∴ C = `pi/2` or A − B = 0
∴ C = `pi/2` or A = B
∴ C = `pi/2` implies that ∆ABC is a right–angled triangle and A = B implies that ∆ABC is an isosceles triangle.
∴ The triangle is either a right–angled triangle or an isosceles triangle.
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