In ∆ABC, if 3∠A = 4 ∠B = 6 ∠C, calculate ∠A, ∠B and ∠C.

Let $3 \angle A=4 \angle B=6 \angle C=x^{\circ}$.

Then,

$\angle A=\left(\frac{x}{3}\right)^{\circ}, \angle B=\left(\frac{x}{4}\right)^{\circ}$ and $\angle C=\left(\frac{x}{6}\right)^{\circ}$

$\therefore \frac{x}{3}+\frac{x}{4}+\frac{x}{6}=180^{\circ} \quad[$ Sum of the angles of a triangle $]$

$\Rightarrow 4 x+3 x+2 x=2160^{\circ}$

$\Rightarrow 9 x=2160^{\circ}$

$\Rightarrow x=240^{\circ}$

Therefore,

$\angle A=\left(\frac{240}{3}\right)^{\circ}=80^{\circ}, \angle B=\left(\frac{240}{4}\right)^{\circ}=60^{\circ}$ and $\angle C=\left(\frac{240}{6}\right)^{\circ}=40^{\circ}$