Question

For a $△ABC;\angle A={45}^{\circ },a=1\text{unit},b=2\text{units}.$ Then possible number of triangle(s) is
(where $a,b,c$ are sides of $△ABC$ opposite to $\angle A,\angle B$ and $\angle C$ respectively)

• A. Zero
• B. One
• C. Two
• D. more than two

Answer 1

The correct option is A Zero

For triangle $ABC:$ using sine rule
$\frac{a}{\sin A}=\frac{b}{\sin B}\Rightarrow \sin B=\frac{2}{1}\cdot \frac{1}{\sqrt{2}}\Rightarrow \sin B=\sqrt{2}>1$
So, there is no such triangle possible.