Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $\angle A={60}^0,a=5,b=2\sqrt{3}$ , then angle $B=?$

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

Answer 1

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

Given, $\angle A={60}^0,a=5,b=2\sqrt{3}$

By sine formula $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$\Rightarrow \frac{5}{\sin {60}^0}=\frac{2\sqrt{3}}{\sin B}\Rightarrow \sin B=\frac{2\sqrt{3}}{5}\times \frac{\sqrt{3}}{2}=\frac{3}{5}$

$\Rightarrow B={\sin }^{-1}\frac{3}{5}$ or ${180}^0-{\sin }^{-1}\frac{3}{5}$ (obtuse angle)

But $a=5,b=2\sqrt{3}\Rightarrow a>b$

$\Rightarrow \angle A>\angle B$

$\Rightarrow \angle B$ must be acute

$\therefore \angle B={\sin }^{-1}\frac{3}{5}$