the plane is 2097.99m away from the tower
Explanation:
We need an illustration to solve this question:
From the diagram, CB is the distance from the tower to the plane
CF = tower's height
BF = plane's height
[tex]\begin{gathered} To\text{ get CB, we will apply sine ratio} \\ \sin \text{ 17.4}\degree=\text{ }\frac{opposite}{hypotenuse} \\ \sin \text{ 17.4}\degree\text{ = }\frac{AC}{CB} \end{gathered}[/tex][tex]\begin{gathered} \sin ce\text{ BF = AC + CF} \\ \text{and BF = 715m, CF = 87.7} \\ AC\text{ = BF - CF = 715 - 87.7} \\ AC\text{ = 627.3 m} \end{gathered}[/tex][tex]\begin{gathered} \sin \text{ 17.4}\degree\text{ = }\frac{627.3}{CB} \\ CB\text{ (sin 17.4}\degree)\text{ = 627.3} \\ CB\text{ = }\frac{627.3}{\sin \text{ 17.4}\degree}\text{ = }\frac{627.3}{0.2990} \\ CB\text{ = }2097.99 \end{gathered}[/tex]
Hence, the plane is 2097.99m away from the tower
