Question

The measure of angles of triangle are $x^0,(x-20{)}^0\text{and}(x-40{)}^0$

.Find the measure of each angle

Answer 1

Sum of angles of a triangle is ${180}^{\circ }$

So,
$x+(x-{20}^{\circ })+(x-{40}^{\circ })$
$\Rightarrow 3x^{\circ }-{60}^{\circ }={180}^{\circ }$

$\Rightarrow 3x^2={240}^{\circ }$

$\therefore x^{\circ }={80}^{\circ }$

So,The angles of the triangle are

$x^{\circ }={80}^{\circ }$

$(x-{20}^{\circ })={60}^{\circ }$

$(x-{40}^{\circ })={40}^{\circ }$