Find the equation of the line passing through `(2,2sqrt(3))` and inclined with x-axis at an angle of `75^0` .

Explanation

Slope of a line inclined `75^(@)` from `x`-axis is, `m=tan75^(@)=tan(45^(@)+30^(@))`
`=(tan45^(@)+tan30^(@))/(1-tan45^(@)tan30^(@))=(1+(1)/(sqrt(3)))/(1-(1)/(sqrt(3)))=(sqrt(3)+1)/(sqrt(3)-1)`
Now, equation of a line of slope `m=(sqrt(3)+1)/(sqrt(3)-1)` and passing through the point `(2,2sqrt(3))` is
`y-2sqrt(3)=(sqrt(3)+1)/(sqrt(3)-1)(x-2)`
`impliesy(sqrt(3)-1)-2sqrt(3)(sqrt(3)-1)`
`=x(sqrt(3)+1)-2(sqrt(3)+1)`
`impliesx(sqrt(3)+1)-y(sqrt(3)-1)-2sqrt(3)-2+6-2sqrt(3)=0`
`implies x(sqrt(3)+1)-y(sqrt(3)-1)+4(1-sqrt(3))=0`