Question
In a $$\vartriangle ABC,a = 2b$$ and $$\left| {A - B} \right| = \frac{\pi }{3}.$$ The measure of $$\angle C$$ is
A.
$$\frac{\pi }{4}$$
B.
$$\frac{\pi }{3}$$
C.
$$\frac{\pi }{6}$$
D.
None of these
Answer :
$$\frac{\pi }{3}$$
Solution :
Clearly, $$A > B\,\,\left( {\because \,\,a > b} \right).$$
Now, $$\tan\frac{{A - B}}{2} = \frac{{a - b}}{{a + b}}\cot \frac{C}{2}$$
$$\eqalign{
& \Rightarrow \,\,\tan {30^ \circ } = \frac{1}{3}\cot \frac{C}{2} \cr
& \therefore \,\,\sqrt 3 = \cot \frac{C}{2}\,\,\,{\text{or,}}\,\,\frac{C}{2} = \frac{\pi }{6}. \cr} $$