The Solution:
Part (a)
Representing the problem fully in a diagram, we have:
Part (b)
We are required to find the length of QA= x.
We shall use Trigonometrical Ratio as below:
[tex]\begin{gathered} tan48^o=\frac{opposite}{adjacent}=\frac{88.9}{x} \\ \\ tan48=\frac{88.9}{x} \end{gathered}[/tex]Making x the subject of the formula, we get
[tex]x=\frac{88.9}{tan48}=80.0459\approx80.0m[/tex]Thus, the distance from the control tower to point A is 80.0 meters.
Part (c)
We are required to find the length of AB= y.
Considering triangle PQB, we have:
[tex]\begin{gathered} tan25.2=\frac{88.9}{x+y}=\frac{88.9}{80+y} \\ \\ 0.47056=\frac{88.9}{80+y} \end{gathered}[/tex]Solving for y, we get
[tex]\begin{gathered} 80+y=\frac{88.9}{0.47056} \\ \\ y=188.922-80 \\ y=108.922\approx108.9m \end{gathered}[/tex]Thus, the distance moved by the plane from its initial position is 108.9 meters.