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
cx tan q
d) x cos q
To illustrate the problem, refer to the following diagram, made with the ever-helpful notepad application:
|\ | \ | \h| \ | \ | \ |_____q\ x
Pretending as if that is a wonderfully-drawn triangle, we are given the distance x of the observer from the launch pad. The angle q is also stated, and is located in the diagram as shown. To look for the distance h of the tip of the rocket to the ground, we consider the following trigonometry function:
tan q = h/x
Since we are solving for h, we multiply both sides with x, giving us:
