Question

The angles of triangle are $x-{40}^o,x-{20}^o$ and $\frac{1}{2}x+{15}^o$. Find the value of x.

Answer 1

Consider the $△ABC$

We know, Sum of angles of a triangle is ${180}^o$ .

Hence $\angle A+\angle B+\angle C={180}^o$

$x-{40}^o+x-{20}^o+\frac{1}{2}x+{15}^o={180}^o$

$\Rightarrow \frac{5}{2}x=(180+45{)}^o={225}^o$

$\Rightarrow x={90}^0$