Question

math 145: mathematics for computing i
in - class activity 5: uncertainty and right angle trigonometry

1. a satellite is directly overhead one observer station when it is at an angle of 68.0° above another observer station. the distance between the two stations is 2000 m. what is the height of the satellite? 2 marks

satellite

1. a small plane takes off from an airport and begins to climb at a 14.5° angle of elevation at 725ft/min.

Explanation:

Step1: Set up the trigonometric relation

Let the height of the satellite be $h$. We have a right - triangle where the angle of elevation is $68.0^{\circ}$ and the adjacent side to the angle is the distance between the two stations, $d = 2000$ m. We know that $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Here, $\theta = 68.0^{\circ}$ and the opposite side is the height of the satellite $h$, and the adjacent side is the distance between the two stations. So, $\tan(68.0^{\circ})=\frac{h}{2000}$.

Step2: Solve for $h$

We can re - arrange the equation $\tan(68.0^{\circ})=\frac{h}{2000}$ to get $h = 2000\times\tan(68.0^{\circ})$. Since $\tan(68.0^{\circ})\approx2.475$, then $h=2000\times2.475 = 4950$ m.

Answer:

$4950$ m