The angle of incline of the hill above the horizontal is 8.81°.
Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.
<h3>Tangent of the angle of the incline of the hill,</h3>
The tangent of the angle of the incline of the hill, Ф is
tanФ = vertical rise/horizontal distance = grade of hill
Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m
So, substituting the values of the variables into the equation, we have
tanФ = vertical rise/horizontal run
tanФ = 15.5 m/100.0 m
tanФ = 0.155
<h3>Angle of incline of the hill</h3>
Taking inverse tan of both sides, we have
Ф = tan⁻¹(0.155)
Ф = 8.81°
So, the angle of incline of the hill above the horizontal is 8.81°.
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