Question

Question 111

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

Answer 1

According to the question,
Measure of $\angle A=\angle B-{40}^{\circ }$
Measure of $\angle C=\angle A+{50}^{\circ }=(\angle B-{40}^{\circ })+{50}^{\circ }$
We know that, the sum of all three angles in a triangle is equal to ${180}^{\circ }$
i.e. $\angle A+\angle B+\angle C={180}^{\circ }$
$\Rightarrow (\angle B-{40}^{\circ })+\angle B+(\angle B-{40}^{\circ }+{50}^{\circ })={180}^{\circ }\Rightarrow 3\angle B-{30}^{\circ }={180}^{\circ }\Rightarrow 3\angle B={210}^{\circ }\therefore \angle B=\frac{{210}^{\circ }}{3}={70}^{\circ }$
So, the measure of $\angle A={70}^{\circ }-{40}^{\circ }={30}^{\circ }$