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:

h = x * tan q

Among the choices, the correct answer is C.

(m+11n)^2=(m+11n)(m+11n)=m^2+11mn+11mn+121n^2=m^2+22mn×121n^2 for example; use foil: (a+b)(a+b)first. a^2outside. abinside. ablast. b^2