Question

If $3x+2y-6=0$ is written in normal form then $\alpha$ is equal to

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

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{2}{3}\right)$

$\alpha \cos \alpha +y\sin \alpha =P$

$3x+2y=6$

$\frac{3x}{\sqrt{13}}+\frac{2y}{\sqrt{13}}=\frac{6}{\sqrt{13}}$

$\frac{3}{\sqrt{13}}x+\frac{2}{\sqrt{13}}y=\frac{6}{\sqrt{13}}$

$\cos \alpha =\frac{3}{\sqrt{13}}$ $\sin \alpha =\frac{2}{\sqrt{13}}$

$\tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{2}{3}$

$\alpha ={\tan }^{-1}\left(\frac{2}{3}\right)$