Question

In a ∆ABC, if $b=\sqrt{3},c=1$ and $B–C=\frac{\mathrm{\pi }}{2},$ then A = ______________.

Answer 1

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

By angle sum property,

Since A + B + C = $\frac{\mathrm{\pi }}{2}$ ....(1)

and by Sine formula,

$$\begin{gathered} \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ \Rightarrow \frac{b}{\sin B}=\frac{c}{\sin C} \\ \Rightarrow \frac{\sqrt{3}}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}+c\right)}=\frac{c=1}{\sin C} \end{gathered}$$
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{\sqrt{3}}{\cos C}=\frac{1}{\sin C}\left(\because \sin \left(\frac{\mathrm{\pi }}{2}+c\right)=\cos C\right) \\ \Rightarrow \frac{\sin C}{\cos C}=\frac{1}{\sqrt{3}} \\ \Rightarrow t\mathrm{an}C=\frac{1}{\sqrt{3}} \\ \mathrm{i}.\mathrm{e}C={30}^{°}=\frac{\mathrm{\pi }}{6} \\ \Rightarrow B=\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{6}=\frac{4\mathrm{\pi }}{6} \\ \Rightarrow A=\mathrm{\pi }-\mathrm{B}-\mathrm{C}(\mathrm{from}(1\left)\right) \\ =\mathrm{\pi }-\frac{2\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6} \\ =\frac{6\mathrm{\pi }-4\mathrm{\pi }-\mathrm{\pi }}{6} \\ \mathrm{i}.\mathrm{e}A=\frac{\mathrm{\pi }}{6} \end{gathered}$$