Question 20.8.3: Reduce the equation √3x+y+2=0 to the normal form x cos α+y s......

We have

Since \cos \alpha<0 and \sin \alpha<0 , \alpha lies in the third quadrant.

Now, \tan \alpha=\frac{\sin \alpha}{\cos \alpha}=\left(\frac{-1}{2}\right) \times\left(\frac{-2}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}=\tan \left(180^{\circ}+30^{\circ}\right)=\tan 210^{\circ} \\  \\ \Rightarrow   \alpha=210^{\circ} .

Thus, \alpha=210^{\circ} and p=1 .

Hence, the given equation in normal form is given by

x \cos 210^{\circ}+y \sin 210^{\circ}=1.