An Airplane Took Off From An Aipory And Traveled At A Constant Rate And Angle Of Elevation. When The Airplane Reached An Altitude Of 500m, Its Horizon

An airplane took off from an aipory and traveled at a constant rate and angle of elevation. When the airplane reached an altitude of 500m, its horizontal distance from the airport was found to be 235m. What was the angle when the airplane rose from the ground?

 

Answer:

The constant angle when the plane rose from the ground is 64.8°.

Step-by-step explanation:

Horizontal Distance along ground from the origin (airport) to the bottom of altitude (height): 235 m

Altitude (height) from the ground: 500 m

Angle of elevation from the origin to the top of altitude: x

Use the Tangent ration to find the angle of elevation:

Tan θ =  opposite/adjacent

Where:

Opposite = Altitude

Adjacent = Horizontal Distance

Tan ⁻¹ θ = (500 m)/(235 m)

Tan ⁻¹ θ = 2.1277

θ = 64.8°

(Significant digits: 3)

The constant angle when the plane rose from the ground is 64.8°.


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