Question

In a $∆\mathrm{ABC},\angle \mathrm{C}=3\angle \mathrm{B}=2(\angle \mathrm{A}+\angle \mathrm{B})$ . Find the three angles.

Answer 1

Step 1: Establish the equations:

Given, $3\angle B=2\left(\angle A+\angle B\right)$

$\Rightarrow$ $3\angle B=2\angle A+2\angle B$

$\Rightarrow$ $\angle B=2\angle A$

Also, $\angle C=3\angle B$

Step 2: Calculate $\angle A$

We know the sum of the angles of a triangle is $180°$.

$\therefore \angle A+\angle B+\angle C=180°$

$\Rightarrow \angle A+\angle B+3\angle B=180$ [Substituting the value of $\angle C$]

$\Rightarrow$ $\angle A+4\angle B=180$

$\Rightarrow \angle A+4\left(2\angle A\right)=180$ [Substituting the value of $\angle B$ ]

$\Rightarrow$ $\angle A+8\angle A=180$

$\Rightarrow$ $9\angle A=180$

$\Rightarrow$ $\angle A=20°$

Step 3: Calculate other angles.

Since, $\angle B=2\angle A$

$$\begin{gathered} \therefore \angle B=20\times 2 \\ =40° \end{gathered}$$

And $\angle C=3\angle B$

$$\begin{gathered} \therefore \angle C=40\times 3 \\ =120° \end{gathered}$$

Hence, the angles of the triangle$\angle A,\angle B$and $\angle C$ are $20°,40°$ and $120°$, respectively.