Question

If the normal form of the equation $\sqrt{3}x+y-8=0$ is $x\cos \omega +y\sin \omega =p$, then $p$ and $\omega$ respectively are

• A. $4,45^{\circ }$
• B. $4,30^{\circ }$
• C. $3,45^{\circ }$
• D. $3,30^{\circ }$

Answer 1

Answer: (B)

Given equation is
$\sqrt{3}x+y-8=\mathrm{0....}$(i)
Dividing Eq. (i) by $\sqrt{(\sqrt{3}{)}^2+(1{)}^2}=2$, we get
$\frac{\sqrt{3}}{2}x+\frac{1}{2}y=4\text{ or }\cos 30^{\circ }x+\sin 30^{\circ }y=\mathrm{4....}$(ii)
Comparing Eq. (ii) with $x\cos \omega +y\sin \omega =p$, we get $p=4$ and $\omega =30^{\circ }$.