Answer
The height of the plane at that time is 1.1 miles
The distance of the plane's path is 10.1 miles
Explanation
The situation models a right angle triangle as shown below
Using trigonometric ratio,
tan 6° = opposite / adjacent
The opposite side of the triangle is the height of the plane. The distance traveled horizontally is the adjacent side. Therefore,
[tex]\begin{gathered} \tan 6^{^0}=\frac{a}{10} \\ 0.1051=\frac{a}{10} \\ \text{Cross multiply} \\ a=10\times0.1051 \\ a=1.051 \end{gathered}[/tex]The height of the plane at that time ≈ 1.1 miles
The hypotenuse of the triangle formed is the distance of the plane path.
Therefore, this distance can be calculated using Pythagoras rule as follows:
[tex]\begin{gathered} c^2=a^2+b^2 \\ a=1.051\text{ miles} \\ b=10\text{ miles} \\ \Rightarrow c^2=1.051^2+10^2 \\ c^2=1.104601+100 \\ c^2=101.104601 \\ c=\sqrt[]{101.104601} \\ c=10.05507837\text{ miles} \end{gathered}[/tex]The distance of the plane's path = 10.1 miles
