Question

A plane leaves the ground with an elevation angle of 6 degrees. The plane travels 10 miles horizontally.How high is the plane at the time?What's the distance of the plane's path?

Answer 1

The height of the plane at that time is 1.1 miles

The distance of the plane's path is 10.1 miles

Step-by-step 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,

`\begin{gathered} \tan 6^(^0)=(a)/(10) \\ 0.1051=(a)/(10) \\ \text{Cross multiply} \\ a=10*0.1051 \\ a=1.051 \end{gathered}`

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:

`\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}`

The distance of the plane's path = 10.1 miles