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# In ΔABC, if a cos A = b cos B, then prove that ΔABC is either a right angled or an isosceles triangle - Mathematics and Statistics

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Sum

In ΔABC, if a cos A = b cos B, then prove that ΔABC is either a right angled or an isosceles triangle

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#### 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.

Concept: Solutions of Triangle
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