Let us consider the situation when the axes are inclined at an angle `omega` . If the coordinates of a point P are `(x_1,y_1)`, then `PN = x_1, PM =y_1` where PM is parallel to the y-axis and PN is parallel to the Xaxis. The straight line through P that makes an angle `theta` with the x-axisis `RQ = y-y_1, PQ = x - x_1 " From " DeltaPQR`, we have `(PQ)/(sin(omega theta))=(sqrtQR)/(sin theta) or y-y_1 = (sin theta)/(sin(omega-theta))(x-x_1)` writen in the form of `y-y_1=m(x-x_1) " where " m=(sin theta)/(sin(omega-theta))` Therefore, if the slope of the line is m, then the angle of inclination of the line with the x-axis is given by an `tan theta=((m sin omega)/(1+m cos omega))`

The axes being inclined at an angle of `60^(@)`, the inclination of the straight line y=2x+5 with the x-axis is

Options

A. `30^(@)`
B. `tan^(-1)(3//2)`
C. `tan^(-1)2`
D. `60^(@)`

Answer

B