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An airplane that had taken off from an airport traveled a grounded distance (horizontal) off 3,660 feet. what is the angle of elevation from the point of takeoff to the point when the plane had traveled 4,150 feet through the air? round to nearest degree.

User Plaureano
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The triangle formed is a right triangle, use the next trigonometric ratio for right triangles to find the measure of angle θ:


\begin{gathered} \cos \theta=(adjacent)/(hypotenuse) \\ \\ \text{adjacent leg: 3660 ft} \\ \text{hypotenuse: 4150ft} \\ \\ \cos \theta=(3660ft)/(4150ft) \\ \\ \theta=\cos ^(-1)((3660ft)/(4150ft)) \\ \\ \theta\approx28 \end{gathered}

Then, the angle of elevation is 28°

An airplane that had taken off from an airport traveled a grounded distance (horizontal-example-1
User Lusha Li
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