Question

The angles of a triangle are $(x-40{)}^{\circ },(x-20{)}^{\circ }$ and ${(\frac{1}{2}x-10)}^{\circ }$. Find the value of $x$.

Answer 1

Since, the sum of three angles of a triangle = ${180}^{\circ }$

$\therefore$ $(x-40{)}^{\circ }+(x-20{)}^{\circ }+{(\frac{1}{2}x-10)}^{\circ }$$={180}^{\circ }$

$\Rightarrow$ $x-{40}^{\circ }+x-{20}^{\circ }+\frac{1}{2}x-{10}^{\circ }={180}^{\circ }$

$\Rightarrow x+x+\frac{1}{2}x-{70}^{\circ }={180}^{\circ }$

$\Rightarrow \frac{2x+2x+1x}{2}={180}^{\circ }+{70}^{\circ }$

$\Rightarrow \frac{5}{2}x={250}^{\circ }$

$\Rightarrow x=\frac{{250}^{\circ }\times 2}{5}$

$\therefore x={100}^{\circ }$

Hence, the value of $x={100}^{\circ }$.