Question

In $△ABC,\angle A=x^0,\angle B=3x^0$ and $\angle C=y^0$. If $3y-5x=30$. Prove that the triangle is right angled.

Answer 1

Sum of angles of a $△ABC$ is equal to ${180}^o$.
$\therefore$ $x+3x+y=180$
$\therefore$ $4x+y=180$ ...........................(1)
Given,
$3y-5x=30$ ................................(2)
Multiplying (1) by 3 we get,
$12x+3y=540$ ...........................(3)
Solving (3) and (2) we get,
$x={30}^o$ and $y={60}^o$
$\therefore$ In $△ABC,\angle A={30}^0,\angle C={60}^0$.

hence$\angle B={180}^0-{30}^0-{60}^0={90}^0$
Since one of the angle measure ${90}^0$ Hence,the triangle is right angled.