# If the cosines of the two of the angles of a triangle are proportional to the oppsite sides show that the trangle is right angled.?

Aug 13, 2018

Such $\triangle$s are isosceles, inclusive of isosceles right angled triangles.

#### Explanation:

$\frac{a}{b} = \sin \frac{A}{\sin} B = \cos \frac{A}{\cos} B$, giving

#sin ( A - B ) = sinA cos B - cos A sin B = 0.So.

$A = B , \mathmr{and} s o , A + B + C = 2 A + C = \pi$

$\Rightarrow$ the$\triangle$ is isosceles that also could be right angled.

Aug 13, 2018

If we consider that the cosines of the two of the angles of a triangle are inversely proportional to the opposite sides then we can get the expected result. Let for $\Delta A B C$

$a \cos A = b \cos B$

$\implies 2 R \sin A \cos A = 2 R \sin B \cos B$

$\implies \sin 2 A = \sin 2 B$

$\implies \sin 2 A = \sin \left(\pi - 2 A\right)$

$\implies 2 A = \pi - 2 A$

$\implies A + B = \frac{\pi}{2}$

Hence $C = \frac{\pi}{2}$. This means the triangle is right angled when cosine of two angles are inversely proportional to opposite sides.