Question

One of the angles of a triangle is ${40}^o$. The difference of other two angles is ${20}^o$. Find the angles.

Answer 1

Let the angles be $\angle A,\angle B,\angle C$

One of the angle be,$\angle A={40}^{\circ }$

Difference between other two is ${20}^{\circ }$

$\Rightarrow \angle B-\angle C={20}^{\circ }-\left(i\right)$

$\therefore \angle A+\angle B+\angle C={180}^{\circ }$ (Sum of all angles of a $△={180}^{\circ }$)

$\Rightarrow {40}^{\circ }+\angle B+\angle C={180}^{\circ }\Rightarrow \angle B+\angle C={140}^{\circ }-\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$

$2\angle B={160}^{\circ }\Rightarrow \angle B={80}^{\circ },\angle C={60}^{\circ }$