Question

ahmed is sitting on a dock watching a float plane in a harbour. At a certain time, the plane is 350 m above the water and 470 m from Ahmed. Determine the angle of elevation of the plane measured from Ahmed, to the nearest degree

Answer 1

Answer:

The angle of elevation of the plane measured from Ahmed is approximately 36.67°

Step-by-step explanation:

The given parameters of the float plane are;

The height of the float plane above the water, y = 350 m

The horizontal distance of the float plane from Ahmed, x = 470 m

Given that Ahmed is sitting on the dock, by the water, by trigonometric ratios, we have;

The height of the float plane, the distance of the plane from Ahmed and the line of sight forming the angle of elevation of the plane measured from Ahmed, form a right triangle

$tan\angle X = \dfrac{Opposite \ leg \ length}{Adjacent \ Leg\ length}$

Therefore

$tan(\theta) = \dfrac{y}{x}$

Where;

θ = The angle of elevation of the plane measured from Ahmed

y = The leg of the right triangle opposite the reference angle

x = The leg  of the right adjacent to reference angle

Therefore;

θ = arctan(y/x) which gives;

θ = arctan(350/470) ≈ 36.67°

The angle of elevation of the plane measured from Ahmed, θ ≈ 36.67°.