a yacht sails.... (Vectors)?

Question

A yacht sails 16 km on a bearing 196° followed by 8 km on a bearing 137°

I would like to calculate the distance (in km to the nearest 0.1 km)and bearing (to the nearest degree) of the yacht's final position from its initial position.

Any help would be greatly appreciated yet again

Answer 1

x1 = (16 km)cos(196) = -15.38018714
y1 = (16 km)sin(196) = -4.410197693

x2 = (8 km)cos(137) = -5.850829613
y2 = (8 km)sin(137) = 5.455986881

x3 = x1 + x2 = -21.2310168
y3 = y1 + y2 = 1.04578919

Distance = sqrt(x3^2 + y3^2) = 21.2567577
Angle = tan^-1(y3 / x3) = -2.819974043 + 180 = 177.180026 degrees

Answer 2

Decompose the vectors as follows:

x(1) = 16*Cos(196)
y(1) = 16*Sin(196)

x(2) = 8*Cos(137)
y(2) = 8*Sin(137)

Now, your final vector has coordinates:
(x(1)+x(2), y(1)+y(2)) = (x(3), y(3))

You can get the reference angle for this from:
Tan(theta) = y(3)/x(3) so
theta = InvTan(y(3)/x(3))

The length of the vector is: sqrt(x(3)^2 + y(3)^2)

And, you get the final angle by looking at what quadrant the coordinates x(3) and y(3) are in and use the reference angle.

Answer 3

This is a (simple) exercise in trigonometry, resolving and adding vectors.

16 x SIN(196) = -4.41 km (west)
16 x COS(196) = -15.38 km (south)

8 x SIN(137) = 5.46 km (east)
8 x COS(137) = -5.85 km (south)

Add considering signs of the terms
1.05 km (east)
21.23 km (south)

Distance (Pythagoras' rule)
21.25 km (southeasterly)

Bearing (tangent rule)
ARCTAN(1.05/21.23) = 180-2.8 = 177.2 degrees.

Answers per problem Distance 21.2 km, Bearing 177 degrees

Answer 4

you have to break down the bearing into vectors (x and y componants ) and then simply add them.

Answer 5

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