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Airplane Vectors, Using Trigonometry

Q: A commuter airplane starts from an airport and takes the route shown in Figure P3.17. It first flies to city A located at 175 km in a direction 30.0° north of east. Next, it flies 150 km 20.0° west of north to city B. Finally, it flies 190 km due west to city C. Find the location of city C relative to the location of the starting point.

A:

You can just use basic Trig a few times.

First, find x1 and y1 (I’m defining these as the x and y components of the first triangle, the one where a is the hypotenuse. I drew them in green in the diagram above).

x1 = 175*cos(30)
y1 = 175*sin(30)

You can do the same thing for the next triangle, where 150 is the hypotenuse. I’ll call the components x2 and y2. (This triangle is orange.)

x2 = 150*sin(20)
y2= 150*cos(20)

(Note that because of the orientation of this triangle, you have the rare case where the horizontal component is sine, and the vertical component is cosine.)

Now I’m going to define a new point, let’s call it Z. This is the point between C and B where the line crosses the y-axis. If you want to find CO, you first need to find CZ and ZO. You now have everything to do this.

CZ = 190 – (x1-x2) [draw some dotted lines across the triangles to see why this is]
CZ = 190 – (175*cos(30) – 150*sin(20))
CZ = 89.75

OZ = y1 + y2
OZ = 175*sin(30) + 150*cos(20)
OZ = 228.45

Now you can find CO by using Pythagorean Theorem.

CO^2 = CZ^2 + OZ^2
CO^2 = 89.75^2 + 228.45^2
CO = 245

Lastly, you need to find the angle. You just use Trig to do this.

Sin(\theta) = 89.75 / 245
\theta = Sin^-1 (89.75/245)
\theta  = 21.5 degrees

So the new position is 245 km away, 21.5 degrees West of North.