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If $m_1, m_2\left(m_1>m_2\right)$ are the slopes of the lines which make an angle of $30^{\circ}$ with the line joining the points $(1,2)$ and $\beta, 4)$, then $\frac{m_1}{m_2}=$
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Verified Answer
The correct answer is:
$7+4 \sqrt{3}$
Slope of $B C=\frac{4-2}{3-1}=1$
Let the slope of $A B=m_1$
$$
\begin{aligned}
\therefore \quad & \tan 30^{\circ}=\left|\frac{m_1-1}{1+m_1}\right| \Rightarrow \pm \frac{1}{\sqrt{3}}=\frac{m_1-1}{1+m_1} \\
& m_1+1= \pm \sqrt{3}\left(m_1-1\right) \\
& m_1+1=\sqrt{3}\left(m_1-1\right) \\
& m_1(\sqrt{3}-1)=\sqrt{3}+1 \\
& m_1=\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}
\end{aligned}
$$
Similarly, we get $m_2=2-\sqrt{3}$
$$
\therefore \quad \frac{m_1}{m_2}=\frac{2+\sqrt{3}}{2-\sqrt{3}}=7+4 \sqrt{3}
$$
Let the slope of $A B=m_1$
$$
\begin{aligned}
\therefore \quad & \tan 30^{\circ}=\left|\frac{m_1-1}{1+m_1}\right| \Rightarrow \pm \frac{1}{\sqrt{3}}=\frac{m_1-1}{1+m_1} \\
& m_1+1= \pm \sqrt{3}\left(m_1-1\right) \\
& m_1+1=\sqrt{3}\left(m_1-1\right) \\
& m_1(\sqrt{3}-1)=\sqrt{3}+1 \\
& m_1=\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}
\end{aligned}
$$
Similarly, we get $m_2=2-\sqrt{3}$
$$
\therefore \quad \frac{m_1}{m_2}=\frac{2+\sqrt{3}}{2-\sqrt{3}}=7+4 \sqrt{3}
$$
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