In a \(\triangle A B C, b: c=\sqrt{3}: \sqrt{2}\) and the angles \(A, B\), \(C\) are in \(\mathrm{AP}\),…
In a \(\triangle A B C, b: c=\sqrt{3}: \sqrt{2}\) and the angles \(A, B\), \(C\) are in \(\mathrm{AP}\), then \(\angle A=\)
\(45^{\circ}\)
\(65^{\circ}\)
\(55^{\circ}\)
\(75^{\circ}\)
Solution
It is given that in a \(\triangle A B C\), angles \(A, B, C\) are in \(A P\), so \(B=60^{\circ}\).
\(\left\{\because A+B+C=180^{\circ}\right\}\)
and it is also given that,
\(\begin{gathered}
\frac{b}{c}=\frac{\sqrt{3}}{\sqrt{2}} \Rightarrow \frac{\sin B}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}} \\
\Rightarrow \quad \frac{\sqrt{3}}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}} \Rightarrow \sin C=\frac{1}{\sqrt{2}} \Rightarrow C=45^{\circ}
\end{gathered}\)
So, angle \(A=180^{\circ}-B-C=180^{\circ}-105^{\circ}=75^{\circ}\)