Question

A television camera at ground level is filming the lift-off of a space shuttle at a point 750 meters from the launch pad. Let θθ be the angle of elevation to the shuttle and let ss be the height of the shuttle.
(a) Write θθ as a function of ss.

(b) Find θθ when s=900meterss=900meters and s=1500meterss=1500meters.

Answer 1

(a)

tanΦ=s/750

Φ=arctan(s/750)

(b)

Φ=arctan(900/750)≈50.19°

Φ=arctan(1500/750)≈63.43°

Answer 2

Answer:

a). $\theta=tan^{-1}(\frac{s}{750} )$

b). θ = 50.19° and θ = 63.43°

Step-by-step explanation:

a). In the figure attached, Space shuttle is at the point A and camera is at point C.

We have to find the expression representing angle θ.

$tan\theta=\frac{AB}{BC}$

$tan\theta=\frac{s}{750}$

$\theta=tan^{-1}(\frac{s}{750} )$

b). we plug in the value s = 900 meters to calculate the value of θ from the given expression.

θ = $tan^{-1}(\frac{900}{750} )$

θ = $tan^{-1}(1.2)$

θ = 50.19°

For s = 1500 meters

$\theta=tan^{-1}(\frac{1500}{750} )$

$\theta=tan^{-1}(2)$

θ = 63.43°