Question

If $b=3,c=4,\angle B=\frac{\pi }{3}$, then the number of triangles that can be constructed is

• A. $0$
• B. $1$
• C. $3$
• D. $2$

Answer 1

The correct option is A $0$

Given $b=3,c=4,\angle B=\frac{\pi }{3}$

We know that as per Sine rule

$\frac{\sin \frac{\pi }{3}}{3}=\frac{\sin C}{4}$

$\frac{\sqrt{3}}{2\cdot 3}=\frac{\sin C}{4}$

$\sin C=2\sqrt{3}$ which is greater than 1

and sin lies between -1 and 1 that means angle C is not possible

Thus Zero triangles can be construct.

Hence A is the correct option