Question

In $∆ABC,\angle C=3\angle B=2(\angle A+\angle B),$ find the measure of each one of $\angle A,\angle B\mathrm{and}\angle C$.

Answer 1

Let $\angle A=x°\mathrm{and}\angle B=y°$
Then, $\angle C=3\angle B=\left(3y\right)°$
Now, we have:
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°$
⇒ x + y + 3y = 180
⇒ x + 4y = 180 ....(i)
Also, $\angle C=2\left(\angle A+\angle B\right)$
⇒ 3y = 2(x + y)
⇒ 2x − y = 0 ....(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 0 ....(iii)
On adding (i) and (iii), we get:
9x = 180 ⇒ x = 20
On substituting x = 20 in (i), we get:
20 + 4y = 180 ⇒ 4y = (180 − 20) = 160 ⇒ y = 40
∴ x = 20 and y = 40
∴ ​$\angle A=20°,\angle B=40°,\angle C=\left(3\times 40\right)°=120°$