Answer:
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
[tex]tan \theta = \frac{Perpendicular}{Base}[/tex]
[tex]tan q^{\circ}= \frac{AB}{BC}[/tex]
[tex]tan q^{\circ}= \frac{AB}{x}[/tex]
[tex]x tan q^{\circ}=AB[/tex]
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
