Question

In a $\triangle A B C$ $a = 2 b$and$| A - B | = \frac { \pi } { 3 }$. The measure of $\angle C$ is

• A. $\frac { \pi } { 4 }$
• B. $\frac { \pi } { 3 }$
• C. $\frac { \pi } { 6 }$
• D. None of these

Answer 1

Answer: (B)

$\frac { \pi } { 3 }$

Answer 2

Clearly, $A > B$ $( \because a > b )$

Now $\tan \frac { A - B } { 2 } = \frac { a - b } { a + b } \cot \frac { C } { 2 }$ ⇒ $\tan 30 ^ { \circ } = \frac { 1 } { 3 } \cot \frac { C } { 2 }$

$\therefore$ $\sqrt { 3 } = \cot \frac { C } { 2 }$ or $\frac { C } { 2 } = \frac { \pi } { 6 }$ ⇒ $C = \frac { \pi } { 3 }$ .