Question

In a triangle $ABC$, the measure of angle $A$ is ${40}^o$ less than the measure of angle $B$ and ${50}^o$ less than that of angle $C$. Find the measure of $\angle A$.

Answer 1

It has been given that,
$\Rightarrow$ $\angle A=\angle B-{40}^o$ ....(1)
$\angle A=\angle C-{50}^o$ ....(2)
Now, in triangle $ABC$,
$\angle A+\angle B+\angle C={180}^o$
$\Rightarrow$ $\angle A+\angle A+{40}^o+\angle A+{50}^o={180}^o$
$\Rightarrow$ $3\angle A={180}^o-{50}^o-{40}^o$
$\Rightarrow$ $3\angle A={90}^o$
$\Rightarrow$ $\angle A={30}^o$