Question

Question 5
In a $\Delta$ ABC, $\angle$C = 3 $\angle$B = 2 ($\angle$A + $\angle$B). Find the three angles.

Answer 1

Given that,
$\angle C=3\angle B=2(\angle A+\angle B)3\angle B=2\angle A+2\angle B\angle B=2\angle A2\angle A-\angle B=0\cdots \left(i\right)$
We know that the sum of the measures of all angles of a triangle is 180°. Therefore,
$\angle A+\angle B+\angle C={180}^{\circ }\angle A+\angle B+3\angle B={180}^{\circ }\angle A+4\angle B={180}^{\circ }\cdots \left(ii\right)$
Multiplying equation (i) by 4, we obtain
$8\angle A-4\angle B=0\cdots \left(iii\right)$
Adding equations (ii) and (iii), we obtain
$9\angle A={180}^{\circ }\angle A={20}^{\circ }$
From equation (ii), we obtain
${20}^{\circ }+4\angle B={180}^{\circ }4\angle B={160}^{\circ }\angle B={40}^{\circ }\angle C=3\angle B=3\times {40}^{\circ }={120}^{\circ }$
Therefore, $\angle A,\angle B,\angle C$ are ${20}^{\circ },{40}^{\circ },and{120}^{\circ }$ respectively.