In a ∆ABC,

Solution:

Given $b=\sqrt{3}, c=1$ and $B-C=\frac{\pi}{2}$ In a triangle $A B C$

By angle sum property, 

Since $A+B+C=\frac{\pi}{2}$   ...(1)

and by Sine formula,

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

$\Rightarrow \frac{b}{\sin B}=\frac{c}{\sin C}$

 

$\Rightarrow \frac{\sqrt{3}}{\sin \left(\frac{\pi}{2}+c\right)}=\frac{c=1}{\sin C}$

i. e $\frac{\sqrt{3}}{\cos C}=\frac{1}{\sin C} \quad\left(\because \sin \left(\frac{\pi}{2}+c\right)=\cos C\right)$

$\Rightarrow \frac{\sin C}{\cos C}=\frac{1}{\sqrt{3}}$

 

$\Rightarrow t \operatorname{an} C=\frac{1}{\sqrt{3}}$

i. e $C=30^{\circ}=\frac{\pi}{6}$

$\Rightarrow B=\frac{\pi}{2}+\frac{\pi}{6}=\frac{4 \pi}{6}$

 

$\Rightarrow A=\pi-B-C$  $($ from $(1))$

$=\pi-\frac{2 \pi}{3}-\frac{\pi}{6}$

 

$=\frac{6 \pi-4 \pi-\pi}{6}$

i.e $A=\frac{\pi}{6}$