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

Given:

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

To do:

 We have to find the value of $x$.

Solution:

We know that,

Sum of the angles in a triangle is $180^o$.

Let the three angles of the triangle be $\angle A=(x - 40)^o, \angle B=(x - 20)^o, \angle C=(\frac{1}{2}x - 10)^o$

Therefore,

$\angle A + \angle B + \angle C = 180^o$

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

$\frac{4x+x}{2}-70^o=180^o$

$\frac{5x}{2}=180^o+70^o$

$x=\frac{2}{5}(250^o)$

$x=100^o$

Hence, the value of $x$ is $100^o$.