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In a ∆ABC, if $\frac{\cos A}{a}=\frac{\cos B}{b}$ show that the triangle is isosceles.


Given: $\frac{\cos \mathrm{A}}{\mathrm{a}}=\frac{\cos \mathrm{B}}{\mathrm{b}}$

Need to prove: $\triangle \mathrm{ABC}$ is isosceles.

$\frac{\cos A}{a}=\frac{\cos B}{b}$

$\Rightarrow \frac{\sqrt{1-\sin ^{2} A}}{a}=\frac{\sqrt{1-\sin ^{2} B}}{b}$

$\Rightarrow \frac{1-\sin ^{2} \mathrm{~A}}{\mathrm{a}^{2}}=\frac{1-\sin ^{2} \mathrm{~B}}{\mathrm{~b}^{2}}$ [Squaring both sides]

$\Rightarrow \frac{1}{a^{2}}-\frac{\sin ^{2} A}{a^{2}}=\frac{1}{b^{2}}-\frac{\sin ^{2} B}{b^{2}}$

We know, $\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}=2 \mathrm{R}$

Therefore, $\frac{\sin ^{2} \mathrm{~A}}{\mathrm{a}^{2}}=\frac{\sin ^{2} \mathrm{~B}}{\mathrm{~b}^{2}}$


$\Rightarrow \frac{1}{a^{2}}=\frac{1}{b^{2}}$

$\Rightarrow a=b$

That means a and b sides are of same length. Therefore, the triangle is isosceles.


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