Question

If the angle $A,B,C$ of a $\Delta ABC$ are in A.P. and

the largest angle is thrice the smallest angle.

Hence find all the angles of the triangle.

Answer 1

Given, angle A, B, C of $\Delta$ABC are in A.P.

Let angle A, B, C be $a-d,a,a+d$ respectively

According to problem,

$3(a+d)=a-d$ ………..$\left(1\right)$

$a-d+a+a+d={180}^o$ (sum of angle of triangle is ${180}^o$)

$3a={180}^o$

$a={60}^o$

Put value of a in equation $\left(1\right)$

$3({60}^o+d)={60}^o-d$

${180}^o+3d={60}^o-d$

$4d=60-{180}^o$

$4d=-{120}^o$

$d=-{30}^o$

$\angle A=a-d={60}^o-(-{30}^o)={90}^o$

$\angle B={60}^o$

$\angle C=a+d={60}^o-{30}^o={30}^o$

$\therefore \angle A={90}^o,\angle B={60}^o,\angle C={30}^o$.