Question

In $△ABC,$ sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively. If $\angle A={60}^{\circ },a=5,b=2\sqrt{3},$ then $\angle B=$

• A. ${\sin }^{-1}\frac{3}{5}$
• B. ${\sin }^{-1}\frac{4}{5}$
• C. ${180}^{\circ }-{\sin }^{-1}\frac{3}{5}$
• D. ${180}^{\circ }-{\sin }^{-1}\frac{4}{5}$

Answer 1

The correct option is A ${\sin }^{-1}\frac{3}{5}$

Given :
$\angle A={60}^{\circ },a=5,b=2\sqrt{3}$
Using sine rule : $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$
$\Rightarrow \frac{5}{\sin {60}^{\circ }}=\frac{2\sqrt{3}}{\sin B}\Rightarrow \sin B=\frac{3}{5}\therefore B={\sin }^{-1}\frac{3}{5}$
or ${180}^{\circ }-{\sin }^{-1}\frac{3}{5}$ (obtuse angle)
As, $a>b$
$\Rightarrow \angle A>\angle B$
So, $\angle B$ is acute angle.
$\therefore \angle B={\sin }^{-1}\frac{3}{5}$