Question

The sum of the base angles of a triangle is ${130}^{\circ }$ and their difference is ${40}^{\circ }.$ Find the measures of all its angles.

Answer 1

Suppose base angles are $B$ and $C$
According to problem:
$\angle B+\angle C={130}^{\circ }...\left(i\right)$ (given)
$\angle B-\angle C={40}^{\circ }($ given $)....(ii)$

Adding eq. (i) and (ii)
$\Rightarrow 2\angle B={170}^{\circ }$
$\Rightarrow \angle B={85}^{\circ }$
$\angle C={130}^{\circ }-{85}^{\circ }={45}^{\circ }$
Hence, $\angle B={85}^{\circ },\angle C={45}^{\circ }$ and $\angle A={50}^{\circ }$