In any triangle ABC if a $cos A = b cos B$ , then the triangle is either isosceles or a right-angled triangle.

True

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False

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Solution

The correct option is A True
For any triangle ,
From cosine rule ,
$cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$

$a cos A = a \frac{b^{2} + c^{2} - a^{2}}{2 b c}$

$cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$

$b cos A = b \frac{a^{2} + c^{2} - b^{2}}{2 a c}$

$⟹ b \frac{a^{2} + c^{2} - b^{2}}{2 a c} = a \frac{b^{2} + c^{2} - a^{2}}{2 b c} a^{2} b^{2} + b^{2} c^{2} - b^{4} = a^{2} b^{2} + a^{2} c^{2} - a^{4} b^{2} c^{2} - b^{4} - a^{2} c^{2} + a^{4} = 0 b^{2} (c^{2} - b^{2}) - a^{2} (c^{2} - a^{2}) = 0 a^{2} (c^{2} - a^{2}) = b^{2} (c^{2} - b^{2}) ⟹ a = b \frac{c^{2} - a^{2}}{b^{2}} = \frac{c^{2} - b^{2}}{a^{2}} ⟹ c^{2} = a^{2} + b^{2}$
So the triangle either must be Isosceles or Right angled

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