Question

Find the equation of a line whose perpendicular distance from the origin is $4units$ and whose normal makes an angle of ${15}^o$ with the positive $x-axis$

Answer 1

Use normal form, $xCos\alpha +ySin\alpha =p⟹\left(1\right)$

$p=4$ and $\alpha =15⁰$

$Cos15⁰=Cos(45⁰-30⁰)$

$=Cos45⁰Cos30⁰+Sin45⁰Sin30⁰$

$=\frac{1}{\sqrt{2}}\frac{3}{\sqrt{2}}+\frac{1}{\sqrt{2}}\frac{1}{2}$

$=\frac{\sqrt{3+1}}{2\sqrt{2}}$

$Sin15⁰=\sqrt{[1-Cos²15⁰]}$

$=\sqrt{\frac{1-\left(\sqrt{3+1}\right)²}{2{\left(\sqrt{2}\right)}^2}}$

$=\sqrt{\frac{\sqrt{(1-(3+1+2)}\sqrt{3}}{8}}$

$=\sqrt{\frac{\sqrt{(8-4-2)}\sqrt{3}}{8}}$

$=\sqrt{\frac{\sqrt{(4-2)}\sqrt{3}}{8}}$

$=\sqrt{\frac{\sqrt{(4-2)}\sqrt{3}}{2\sqrt{2}}}$

put $Sin15⁰,Cos15⁰,p=4$ in equation (1)

$x\left[Cos15\right]+y\left[Sin15\right]=4$

$x\left[\right(\sqrt{3+1}\left)\right]+y[4-2\sqrt{3}]=4\ast 2\sqrt{2}$

$x\left[\right(\sqrt{3+1}\left)\right]+y[4-2\sqrt{3}]=8\sqrt{2}$