Question

If in a triangle $ABC,b=\sqrt{3},c=1$ and $B-C={90}^o$ then $\angle A$ is

• A. ${30}^o$
• B. ${45}^o$
• C. ${75}^o$
• D. ${90}^o$

Answer 1

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

$Formulaused$
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
$Given$
$b=\sqrt{3},c=1,B-C=90$
$Ina\Delta ABC$
$A+B+C=180$
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
$\frac{\sqrt{3}}{\sin (90+C)}=\frac{1}{\sin C}$
$\Rightarrow \frac{\sqrt{3}}{\cos C}=\left(\frac{1}{\sin C}\right)$
$\Rightarrow \tan C=\frac{1}{\sqrt{3}}=\tan {30}^0$
$HenceC={30}^0$
$B={90}^0+{30}^0={120}^0$
$A+B+C={180}^0$
$A={30}^0$