An observer on the ground is x meters from the base of the launch pad of a rocket, which is at the same level as the observer. a few seconds after the rocket takes off vertically, the observer sees its tip at an angle of q° from the horizontal. how far above the ground is the tip of the rocket at that instant? assume that the ground is level.
a. x/tan q
b. x/ sin q
c. x tan q
d. x sin q
e. x cos q
pls 30 points

x tan q

Step-by-step explanation:

Refer the attached figure

An observer on the ground is x meters from the base of the launch pad of a rocket i.e. BC = x

The height of the rocket is AB

The observer sees its tip at an angle of q° from the horizontal.i.e.∠ACB = q°

In ΔABC

We will use trigonometric ratio

So, the height of the rocket is x tan q.

So, Option C is correct.

Hence the tip of the rocket is  x tan q far above the ground

is there a picture

step-by-step explanation:

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)