Question

The angles of a triangle are (x − 40) , (x − 20) and (1/2 x − 10) . Find the value of x.
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Answer 1

Given:-

• Angles of triangle are (x - 40°) , (x - 20°) and (1/2x - 10°).

To find:

• Value of x

SolutioN:-

We know that,

Sum of all interior angles of triangle is 180°.

So,According to question

$\implies \sf\: (x - 40\degree) + (x - 20\degree) + ( \frac{1}{2} x - 10\degree) = 180\degree$

$\implies \sf\: x - 40\degree + x - 20\degree + \frac{1}{2} x - 10\degree = 180\degree$

$\implies \sf\: x + x + \frac{1}{2} x + 40\degree - 20\degree - 10\degree = 180\degree$

$\implies \sf\: x + x + \frac{1}{2} x + 40\degree - 20\degree - 10\degree= 180\degree$

$\implies \sf\: 2x + \frac{1}{2} + 10\degree= 180\degree$

$\implies \sf\: \frac{4x + x}{2} = 180\degree - 10\degree$

$\implies \sf\: \frac{5x}{2} = 170\degree$

$\implies \sf\: 5x = 170\degree \times 2$

$\implies \sf\: 5x = 340\degree$

$\implies \sf\: x = \frac{340\degree}{5}$

$\implies \sf\: x = 68\degree$

Verification:-

$\implies \sf\: (x - 40\degree) + (x - 20\degree) + ( \frac{1}{2} x - 10\degree) = 180\degree$

• Put x = 68

$\implies \sf\: (68\degree - 40\degree) + (68\degree - 20\degree) + ( \frac{1}{2} \times 68\degree - 10\degree) = 180\degree$

$\implies \sf\: 108\degree + 48\degree + (34\degree - 10\degree)= 180\degree$

$\implies \sf\: 108\degree + 48\degree + 24\degree= 180\degree$

$\implies \sf\: 180\degree = 180\degree$

Therefore,

Value of x is 68°.