a maths

Let the angle of elevation be A
Since the road is inclined at anangle of 30 degree, so the angle between the road surface and the top of the pole = ( A - 30).
Let length of shadow = y.
so 18/y = tan( A - 30)................(1)
-18/y^2 (dy/dt) = [1/cos^2(A - 30)][dA/dt].
Now dA/dt = 1/4 degree per min.
A = 60 degree.
From (1), y = 18/tan 30., cos(A- 30) = cos 30.
so dy/dt = [1/cos^2 30][1/4][18/tan 30]^2[-1/18] = -(4/3)(1/4)(18)/(3)
= -18 m/min.