Question

In a $\Delta ABC,\angle A=x^{\circ },\angle B=3x^{\circ }$ and $\angle C=y^{\circ }.$ If $5x-3y+30=0,$ then prove that it is a right angled.

Answer 1

Since it is given that
$\angle A=x^{\circ },\angle B=3x^{\circ }$ and $\angle C=y^{\circ }.$
$\angle A+\angle B+\angle C={180}^{\circ }$ (by angle sum property of a triangle)
$\Rightarrow x+3x+y={180}^{\circ }$
$\Rightarrow 4x+y={180}^{\circ }$ .......(i)
Also $5x-3y+30=0$
$\Rightarrow 5x-3y=-30$ ......(ii)
Multiply equation (i) by $3$ to make coefficient of y equal

$12x+3y=540$

and then adding with (ii), we get
$17x=510\Rightarrow x={30}^{\circ }$
Now putting the value of x in equation (i), we get
$4\times {30}^{\circ }+y-{180}^{\circ }$
or $y={180}^{\circ }-{120}^{\circ }$
or $y={60}^{\circ }$
Sum of two angles $x$ and $y$ is ${30}^o+{60}^o={90}^{\circ }$
Hence, it is a right angled triangle.