Question

If the angle between two lines is $\frac{\pi }{6}$ and slope of one of the lines is $\frac{1}{2}$, then the slope of the other line can be

• A. $\frac{\sqrt{3}-2}{2\sqrt{3}-2}$
• B. $\frac{\sqrt{3}+2}{2\sqrt{3}-1}$
• C. $\frac{\sqrt{3}-2}{2\sqrt{3}+1}$
• D. $\frac{\sqrt{3}+2}{2\sqrt{3}+1}$

Answer 1

Answer: (B), (C)

$\frac{\sqrt{3}-2}{2\sqrt{3}+1}$

Let slope of the lines be $m_1$ and $m_2$.
Then $m_1=\frac{1}{2}$ and $\theta =\frac{\pi }{6}$
$\therefore \tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$
$\Rightarrow \tan \frac{\pi }{6}=\left|\frac{m_2-\frac{1}{2}}{1+\frac{m_2}{2}}\right|$
$\Rightarrow \frac{1}{\sqrt{3}}=\pm \frac{m_2-\frac{1}{2}}{1+\frac{m_2}{2}}$

Case $\left(1\right)$
When $\frac{1}{\sqrt{3}}=\frac{m_2-\frac{1}{2}}{1+\frac{m_2}{2}}$
$\Rightarrow m_2+2=2\sqrt{3}{m}_2-\sqrt{3}\Rightarrow m_2=\frac{2+\sqrt{3}}{2\sqrt{3}-1}$

Case $\left(2\right)$
When $\frac{1}{\sqrt{3}}=-\frac{m_2-\frac{1}{2}}{1+\frac{m_2}{2}}$
$\Rightarrow m_2+2=-2\sqrt{3}{m}_2+\sqrt{3}\Rightarrow m_2=\frac{\sqrt{3}-2}{2\sqrt{3}+1}$