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?
 A: 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...
A: 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.
A: On putting value of $x$ in $\angle C$ we get,
$\angle C = 32.72$
As we are not getting exact value as in options. So we approximate it near to options I think option (4) is correct.
A: we have $$A=t,2B=t,3C=t$$ thus we get $$t+\frac{t}{2}+\frac{t}{3}=180^{\circ}$$
and we obtain $$11t=180^{\circ}\cdot 6$$
A: As A=2B=3C,
B=1.5C
Therefore, summing the angles, A+B+C= 3C+1.5C+C=180 degrees.
5.5C=180  
C=36/1.1
  =360/11
  = 32.77 degrees. 
