Find the angle between the lines $2y – \sqrt{3}x – 5 = 0$ and $\sqrt{3}y-2x+6=0$.
I do know the formula for finding the angle $\alpha$ between two lines which is
$$
\tan \alpha = \frac{m_2-m_1}{1+m_1m_2}
$$
but I am trying to find the value of the fraction above and I am getting the value of $\frac{\sqrt{3}}{3}$ which isn't any value of any degree for $\tan \alpha$. Am I doing something wrong?
Best Answer
$$m_1=\frac{\sqrt3}{2}$$ $$m_2=\frac{2}{\sqrt3}$$
Thus the angle between line 1 and line 2
$$\tan\alpha=\frac{\frac{2}{\sqrt3}-\frac{\sqrt3}{2}}{1+\frac{2}{\sqrt3}\frac{\sqrt3}{2}}$$ $$\tan\alpha=\frac{1}{4\sqrt3}$$
Thus the angle $\alpha$ is $\arctan(\frac{1}{4\sqrt3})\approx 8.213^\circ.$