In a $\triangle ABC, \angle A = x^o, \angle B = 3x^o and \angle C = y^o$. If $3y - 5x = 30$, prove that the triangle is right angled.

Given:

In a $\triangle ABC, \angle A = x^o, \angle B = 3x^o and \angle C = y^o$.

$3y - 5x = 30$

To do:

We have to prove that the triangle is right angled.

Solution:

We know that,

Sum of the angles in a triangle is $180^o$. 

Therefore,

$\angle A+\angle B+\angle C=180^o$

$x^o+3x^o+y^o=180^o$

$4x^o+y^o=180^o$

$y^o=180^o-4x^o$....(i)

$3y - 5x = 30$

$3(180-4x)-5x=30$    (From (i))

$540-12x-5x=30$

$17x=540-30$   

$17x=510$

$x=\frac{510}{17}$

$x=30$

$y=180-4(30)$     (From (i))

$y=180-120$

$y=60$

This implies,

$\angle A = x^o=30^o$

$\angle B = 3x^o=3(30^o)=90^o$

$\angle C = y^o=60^o$

Therefore, $\triangle ABC$ is a right-angled triangle with the right angle at B. 

Hence proved.

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Updated on: 10-Oct-2022

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