Question

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

Answer 1

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.