Question

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

Answer 1

Let $\angle A=x^o,\angle B=y^o$. Then,

$\angle C=3\angle B\Rightarrow \angle C=3y^o$

$\therefore 3\angle B=2(\angle A+\angle B)$

$\Rightarrow 3y=2(x+y)\Rightarrow y=2x\Rightarrow 2x-y=0$ .(i)

Since $\angle A,\angle B,\angle C$ are angles of a triangle.

$\therefore \angle A+\angle B+\angle C={180}^o$

$\Rightarrow x+y+3y=180\Rightarrow x+4y=180$ ..(ii)

Putting $y=2x$ in equation (ii), we get
$x+8x=180\Rightarrow 9x=180\Rightarrow x={20}^o$

Putting the value of x in equation (i), we get $y={40}^o$

Hence, $\angle A={20}^o,\angle B={40}^o$ and $\angle C=3y^o=(3\times {40}^o)={120}^o$.