In a $\triangle A B C$, if $b=\sqrt{3}, c=1$ and $B-C=\frac{\pi}{2}$, then $A=$
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 $A=\frac{\pi}{6}$
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