Help me with this calculus problem. This is a related rates problem. Do not use inverse trigonometric functions.

Question

A man whose eye level is 6 ft above the ground walks toward a billboard at a rate of 2 ft/s. The bottom of the billboard is 10 ft above the ground and it is 15 ft high. The man's viewing angle is the angle formed by the lines between the man's eyes and the top and bottom of the billboard. At what rate is the viewing angle changing when the man is 30 ft from the billboard?

Answer 1

Method I. Use inverse trig.

Let x = the horizontal distance from the man to the billboard, θ = the viewing angle.

θ = arctan(19/x) - arctan(4/x),

Differentiate both sides with respect to time, and plug in x = 30, x' = -2

θ' = 38/(30^2+19^2) - 8/(30^2+4^2) = 0.0214 rad/sec <==Answer

 

Method II. Use law of cosines,

15^2 = (x^2+4^2) + (x^2+19^2) - 2sqrt[(x^2+4^2) (x^2+19^2)] * cos θ

Can you finish it?