<s>[INST] 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?
 [/INST] It looks like neither of the options is correct. $$\angle C = \frac{x}{3} = \frac{180 \times 6}{11 \times 3} = \frac{180 \times 2}{11} = \frac{360}{11} = 32 \frac{9}{11}  $$
You can also calculate $A+B+C = 3C + \frac{3}{2}C + C = \frac{11}{2} C$ for all possibile answers to show that none of them gives $180^\circ$ for the sum of the angles $A+B+C$, and hence none of them is correct.
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