Question

In a triangle ABC if $BC=1$ and $AC=2$ ,then what is the maximum possible value of angle A.

• A. ${\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$
• B. ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• C. ${\tan }^{-1}\left(\frac{1}{2}\right)$
• D. ${\tan }^{-1}\left(\frac{1}{3}\right)$

Answer 1

Answer: (B)

${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Given $BC=1,AC=2$

using sine rule

$\frac{1}{\sin A}=\frac{2}{\sin B}$ or $\sin A=\frac{\sin B}{2}$

And, maximum value of $\sin B=1$

$\Rightarrow$ Maximum value of $\sin A=\frac{1}{2}$

$\cos A=\sqrt{1-{\sin }^{2}A}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

$\tan A=\frac{1}{\sqrt{3}}$

$\Rightarrow A={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$