Question

If in triangle ABC ,cosA/a=cosB/b ,prove that triangle is isosceles

Answer 1

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Answer 2

Answer:

Given,

In a triangle ABC ( shown below ),

$\frac{cos A}{a}=\frac{cos B}{b}$ ------(1)

We know that,

By the law of sine,

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

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

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

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

From equation (1),

$\frac{1}{a^2}-\frac{cos^2B}{b^2}=\frac{1}{b^2}-\frac{cos^2B}{b^2}$

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

$\implies a = b$

$\implies AC = BC$

Hence, triangle ABC is an isosceles triangle.