Question

In a triangle $ABC,b=\sqrt{3},c=1$ and $\angle A={30}^o$, then the largest angle of the triangle is

• A. ${135}^o$
• B. ${90}^o$
• C. ${60}^o$
• D. ${120}^o$

Answer 1

The correct option is B ${90}^o$

$\begin{array}{c}In,\Delta ABC\\ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\ \Rightarrow \frac{a}{\sin {30}^o}=\frac{\sqrt{3}}{\sin B}=\frac{1}{\sin C}\\ \Rightarrow \frac{a}{1/2}=\frac{\sqrt{3}}{\sin B}\\ \Rightarrow 2a=\frac{\sqrt{3}}{\sin B}\\ \Rightarrow \sin B=\frac{\sqrt{3}}{2a}=\frac{\sqrt{3}}{2}=\sin {60}^o\\ \Rightarrow B={60}^o\\ cosA=\frac{b^2+c^2-a^2}{2bc}\\ \Rightarrow cos{30}^o=\frac{{\left(\sqrt{3}\right)}^2+1^2-a^2}{2.\sqrt{3}.1}=\frac{4-a^2}{2\sqrt{3}}\\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{4-a^2}{2\sqrt{3}}\\ \Rightarrow 3=4-a^2\Rightarrow a=1\\ Now,\\ A+B+C={180}^o\\ \Rightarrow {30}^o+{60}^o+C={180}^o\\ \Rightarrow C={90}^o\end{array}$