Question

In $\Delta ABC$, $2\angle A=3\angle B=6\angle C$. Find $\frac{\angle A+\angle B}{\angle C}\times \angle B$

Answer 1

$2\angle A=3\angle B=6\angle C$

So, we can write

$\angle A=3\angle C\angle B=2\angle C$

and in a Triangle Sum of all interior angles is ${180}^o$

$\angle A+\angle B+\angle C={180}^o\Rightarrow 6\angle C={180}^o\Rightarrow \angle C={30}^o$

Now,

$\frac{\angle A+\angle B}{\angle C}.\angle B=\frac{3\angle C+2\angle C}{\angle C}.2\angle C=10\angle C=10\ast {30}^o={300}^o$