Captain Jack Sparrow has been marooned on an island in the Atlantic by his crew, and decides to builda raft to escape. The wind seems quite steady, and first blows him due east for 11km, and then 6km ina direction 6degrees north of east. Confident that he will eventually find himself in safety, he fallsasleep. When he wakes up, he notices the wind is now blowing him gently 11degrees south of east -but after traveling for 21km, he finds himself back on the island.

Variable Name Min Max Step Sample Value
thetab 5 10 1 6
a 10 20 11 1
b 5 15 1 6
c 20 30 1 21
thetac 10 15 11 1

Required:
How far (in km) did the wind blow him while he was sleeping?

Answer:

    d₃ = 37,729 km,     θ=  5.1º North of West

Explanation:

This is a velocity addition problem, the easiest way to solve it is to decompose the velocities in a Cartesian system, the x-axis coincides with the West-East direction and the y-axis with the South-North direction

* first displacement is

           d₁ₓ = 11 km

* second offset is

          cos 6 = d₂ₓ / d₂

          sin 6 = d_{2y} / d₂

          d₂ₓ = d₂ cos 6

          d_{2y} = d₂ sin 6

          d₂ₓ = 6 cos 6 = 5.967 km

          d_{2y} = 6 sin 6 = 0.6272 km

* third displacement is unknown

* fourth and last displacement

          cos (-11) = d₄ₓ / d₄

          sin (-11) = d_{4y} / d₄

          d₄ₓ = d₄ cos (-11)

          d_{4y} = d₄ sin (-11)

          d₄ₓ = 21 cos (-11) = 20.61 km

          d_{4y} = 21 sin (-11) = -4.007 km

They tell us that at the end of the tour you are back on the island, so the displacement must be zero

X axis

           x = d₁ₓ + d₂ₓ + d₃ₓ + d₄ₓ

           0 = 11 +5.967 + d₃ₓ + 20.61

           d₃ₓ = -11 - 5.967 - 20.61

           d₃ₓ = -37.577 km

Y axis  

          y = d_{1y} + d_{2y} + d_{3y} + d_{4y}

          0 = 0 + 0.6272 + d_{3y} -4.007

          d_{3y} = 4.007 - 0.6272

          d_{3y} = 3.3798 km

This distance can be given in the form of module and angle

Let's use the Pythagorean theorem for the module

           d₃ = [tex]\sqrt{d_{3x}^2 + d_{3y}^2}[/tex]

           d₃ = [tex]\sqrt{37.577^2 + 3.3798^2}[/tex]

           d₃ = 37,729 km

Let's use trigonometry for the angle

            tan θ = d_{3y} / d₃ₓ

            θ = tan⁻¹ [tex]\frac{d_{3y}}{d_{3x}}[/tex]

            θ = tan-1 (-3.3798 / 37.577)

            θ = 5.1º

Since the y coordinate is positive and the x coordinate is negative, this angle is in the second quadrant, so the direction given in the form of cardinal coordinates is

            θ=  5.1º North of West

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