Help me, please!!! Thank you so much!

Question

A SpaceX rocket has launched from Virginia, and I am watching it from5 miles away. When the rocket is 8 miles in the air(assuming it goes straight up, which it doesn’t really ), what is the angle of elevation I am looking at it with?

Answer 1

Since we're assuming the rocket traveled straight up, we have a 90° angle.

We have x = 5, y = 8, so distance between the rocket and you is d² = 5² + 8² = 89, or d = √(89)

Using law of sines, we get

√(89)/sin(90) = 8/sin(θ) or sin(θ) = 8/(√(89)) ⇒ θ ≈ 58°

Answer 2

Hmmm. Are you taking the ground as "flat" or curved like the earth's surface? I am presuming you are assuming the ground is "flat", (earth's curvature not taken into account).

If so, then you would just have a right triangle with the vertical leg at 8 mi and the base leg is 5 mi. To calculate the angle of elevation, you would use Tan θ = 8/5; so θ = Tan^-1 (8/5), you will find θ = 57.99 degrees, or 57 degrees, 59 minutes, 41 seconds.

Answer 3

The straight up path of the rocket (8 miles) and your horizontal distance from the rocket (5 miles) form the legs of a right triangle.

The angle of elevation is adjacent to the 5 miles side and opposite to the 8 mile side, so

tan (angle) = opposite/adjacent = 8/5 = 1.6

angle = arctan(1.6) = 58 degrees.