In a triangle ABC,if a cos A = b cos B,where | vedclass

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(D) Given $a \cos A = b \cos B$. Using the cosine rule $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$: $a \left(\fr

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Right angled triangle

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(D) Given $a \cos A = b \cos B$. Using the cosine rule $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$: $a \left(\frac{b^2 + c^2 - a^2}{2bc}ight) = b \left(\frac{a^2 + c^2 - b^2}{2ac}ight)$ $\Rightarrow \frac{a}{b} (b^2 + c^2 - a^2) = \frac{b}{a} (a^2 + c^2 - b^2)$ $\Rightarrow a^2 (b^2 + c^2 - a^2) = b^2 (a^2 + c^2 - b^2)$ $\Rightarrow a^2b^2 + a^2c^2 - a^4 = a^2b^2 + b^2c^2 - b^4$ $\Rightarrow a^4 - b^4 + b^2c^2 - a^2c^2 = 0$ $\Rightarrow (a^2 - b^2)(a^2 + b^2) - c^2(a^2 - b^2) = 0$ $\Rightarrow (a^2 - b^2)(a^2 + b^2 - c^2) = 0$ Since $a 
eq b$,we have $a^2 - b^2 
eq 0$. Therefore,$a^2 + b^2 - c^2 = 0$,which implies $a^2 + b^2 = c^2$. Thus,$	riangle ABC$ is a right-angled triangle.

In a $\triangle ABC$,if $a \cos A = b \cos B$,where $a \neq b$,then $\triangle ABC$ is

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