Question
In $\triangle \mathrm{ABC}$ if $\frac{\cos A}{a}=\frac{\cos B}{b}$, then show that it is an isosceles triangle.
✓
Answer
Given: $\frac{\cos A}{a}=\frac{\cos B}{b} \ldots(1)$
By sine rule,
$\frac{a}{\sin A}=\frac{b}{\sin B}=k$
$\therefore a=k \sin A, b=k \sin B$
$\therefore$ (1) gives, $\frac{\cos A}{k \sin A}=\frac{\cos B}{k \sin B}$
$\therefore \frac{\cos A}{\sin A}=\frac{\cos B}{\sin B}$
∴ sin A cos B = cos A sinB ∴ sinA cosB – cosA sinB = 0 ∴ sin (A – B) = 0 = sin0 ∴ A – B = 0 ∴ A = B ∴ the triangle is an isosceles triangle.
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