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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 $100^o$.
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