Answer to Question #96582 in Calculus for Amir aziz

Question

A helicopter takes off from a point 80 m away from an observer located on theground, and rises vertically at 2 m/s. At what rate is elevation angle of theobserver’s line of sight to the helicopter changing when the helicopter is 60 mabove the ground In the answer of this question there is (when t=30 cos=4/5) i did not got this point sir can you please explain it .it will really help me

Answer 1

Let's put observer in point A and helicopter in point B (on the ground). $\alpha$ is the elevation angle of the observer’s line of sight to the helicopter.

$BC = 2\cdot t$ m because the helicopter rises by 2 meters every second. By the time of observation, $BC = 2\cdot t =60$ m, thus, 30 seconds have passed ($t = 30$ s).

By the Pythagorean theorem applied to the right triangle ABC,

$AB^2+BC^2=AC^2 \\$

$AC = \sqrt{80^2+(2t)^2} = \sqrt{80^2+60^2} =\sqrt{10000} = 100$ m.

By the definition of the cosine function,

$cos{\alpha}= \dfrac{AB}{AC} = \dfrac{80}{100} = \dfrac{4}{5}$