Find the equation of the straight line which makes an angle of `15^@ ` with the positive direction of x-axis and which cuts and intercept of length `4` on then negative direction of y-axis.

Explanation

Here , m = slope of the line = `tan15` = `(sqrt3-1)/(sqrt3+1) =(sqrt3-1)/(sqrt3+1) * (sqrt3-1)/(sqrt3-1) = (2-sqrt3)`


and c= intercept on y - axis = -4


Therefore, equation of the required line will be `y = (2-sqrt3) x -4 `.
or


`(2-sqrt3)x-y-4=0`