Question

For a $△ABC;\angle B={60}^{\circ },b=3,c=4.$ Then possible number of triangle(s) is
(where $a,b,c$ are sides(in $\text{units}$) of $△ABC$ opposite to $\angle A,\angle B$ and $\angle C$ respectively)

• A. Zero
• B. One
• C. Two
• D. Infinite

Answer 1

The correct option is A Zero

Given : in $△ABC;\angle B={60}^{\circ },b=3,c=4.$
Using sine rule :
$\Rightarrow \frac{\sin C}{\sin B}=\frac{c}{b}\Rightarrow \sin C=\frac{4}{3}\cdot \frac{\sqrt{3}}{2}\Rightarrow \sin C=\frac{2}{\sqrt{3}}>1$
So, there is no such triangle possible.