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:
![\begin{gathered} tan48^o=(opposite)/(adjacent)=(88.9)/(x) \\ \\ tan48=(88.9)/(x) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/j52tzq56oro0f13isr6p2mzv6302u43jy2.png)
Making x the subject of the formula, we get
![x=(88.9)/(tan48)=80.0459\approx80.0m](https://img.qammunity.org/2023/formulas/mathematics/college/2l3h716wudj237a6km7zd1t5c549fukmk1.png)
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:
![\begin{gathered} tan25.2=(88.9)/(x+y)=(88.9)/(80+y) \\ \\ 0.47056=(88.9)/(80+y) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gds8ti13jhw79fmp10zdmm8dmzgdjqsn8p.png)
Solving for y, we get
![\begin{gathered} 80+y=(88.9)/(0.47056) \\ \\ y=188.922-80 \\ y=108.922\approx108.9m \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/eymr1vxdqmuz5ay46rdzzr7iyjnrk2r3p5.png)
Thus, the distance moved by the plane from its initial position is 108.9 meters.