In a triangle $ABC$, if its angles are such that $A=2B=3C$ then find $\angle C$? In a triangle $ABC$, if its angles are such that $A=2B=3C$ then find $\angle C$?
$1$. $18°$
$2$. $54°$
$3$. $60°$
$4$. $30°$
My Attempt:
$$A=2B=3C$$
Let $\angle A=x$ then $\angle B=\dfrac {x}{2}$ and $\angle C=\dfrac {x}{3}$
Now, 
$$x+\dfrac {x}{2} +\dfrac {x}{3}=180°$$
$$\dfrac {6x+3x+2x}{6}=180°$$
$$\dfrac {11x}{6}=180°$$
$$x=\dfrac {180\times 6}{11}$$
So, which is the correct option?

Add:
$$180^\circ=A+B+C=2B+B+\frac23B=\frac{11}3B\implies B=\frac{540}{11},\ldots etc.$$
Or also
$$180^\circ=A+B+C=3C+\frac32C+C=\frac{11}2C\implies C=\frac{360}{11}$$
By the way, none of the options given in your question matches the above...
