MHT CET · Maths · Properties of Triangles
In a triangle \(A B C\), with usual notations, if \(b=\sqrt{3}, c=1, \angle A=30^{\circ}\), then angle \(B\) is
- A \(60^{\circ}\)
- B \(90^{\circ}\)
- C \(30^{\circ}\)
- D \(120^{\circ}\)
Answer & Solution
Correct Answer
(D) \(120^{\circ}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \cos A=\frac{b^2+c^2-a^2}{2 b c} \\ & \Rightarrow \cos 30^{\circ}=\frac{3+1-a^2}{2 \times \sqrt{3} \times 1} \\ & \Rightarrow \frac{\sqrt{3}}{2}=\frac{4-a^2}{2 \sqrt{3}} \\ & \Rightarrow a=1\end{aligned}\)
Now \(\cos B=\frac{c^2+a^2-b^2}{2 c a}=\frac{1^2+1^2-(\sqrt{3})^2}{2 \times 1 \times 1}=-\frac{1}{2}\)
\(\Rightarrow B=120^{\circ}\)
Now \(\cos B=\frac{c^2+a^2-b^2}{2 c a}=\frac{1^2+1^2-(\sqrt{3})^2}{2 \times 1 \times 1}=-\frac{1}{2}\)
\(\Rightarrow B=120^{\circ}\)
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