Question

In an acute angled $\Delta ABC$, $a=4$ cm, $b=6$ cm, $\sin B=\frac{3}{4}$, then the value of $\angle A$ is

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. None of these

Answer 1

The correct option is A ${30}^0$

It is given that in an acute angled triangle ABC $a=4cm,b=6cm,\sin B=\frac{3}{4}$

applying formula for \sin es we get

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

Therefore

$\frac{a}{\sin A}=\frac{b}{\sin B}$

Substituting the values of $a,b,\sin B$ we get

$\frac{4}{\sin A}=\frac{6}{\frac{3}{4}}$

$\frac{4}{\sin A}=\frac{4\left(6\right)}{3}$

$\sin A=\frac{3\left(4\right)}{4\left(6\right)}$

$\sin A=\frac{1}{2}$

Hence $A={30}^0$