Question

In $\Delta$ABC, $\angle C=2\angle B=\angle A+\angle B+{20}^o$. Find all the angles of the triangle.

Answer 1

Let $\Delta$ABC, $\angle A=x^o$ and $\angle B=y^o$
We know that $\angle A+\angle B+\angle C={180}^o$
Then, $x^o+y^o+\angle C={180}^o$
or $\angle C={180}^o-(x^o+y^o)$
Given $\angle C=2\angle B$
${180}^o-(x^o+y^o)=2y^o$
$x^o+3y^o={180}^o$ ..........(i)
Again given
$2\angle B=\angle A+\angle B+20$
$2y^o=x^o+y^o+20$
$x^o-y^o=-20$ ............(ii)
Solving equation (i) and (ii)
We get $x={30}^o,y={50}^o$
$\therefore \angle A=x^o={30}^o$
$\angle B=y^o={50}^o$
$\angle C=2\angle B=2\times {50}^o={100}^o$.