Question

In a $△$ ABC, if a cos A = b cos B, show that the triangle is either isosceles or right angled.

Answer 1

Using sine rule, $\frac{a}{\text{sin A}}=\frac{b}{\text{sin B}}=\frac{c}{\text{sin C}}$ = k

Then,

Now, acos A = bcos B

$\Rightarrow$ k sin A cos A = k sin B cos B

$\Rightarrow$ 2 sin A cos A = 2 sin B cos B [multiply by 2]

$\Rightarrow$ sin 2 A = sin 2 B

$\Rightarrow$ 2 A = 2 B or 2 A = $\pi$ - 2 B

$\Rightarrow$ A = B or A + B = $\frac{\pi }{2}$

$\Rightarrow$ A = B or C = $\frac{\pi }{2}$

Hence, $△$ ABC is either isosceles or right angled.

Hence proved.