<p>An airplane flies 200 km due west from city A to city B and then <span style=

Question

<p>An airplane flies 200 km due west from city A to city B and then <span>245</span> km in the direction of <span>25.0</span><span>&#176;</span> north of west from city B to city C.</p>
<p>Relative to city A, in what direction is city c?&#160;&#160; degrees north of west</p>
<p>&#160;</p>
<p>&#65279;&#65279;&#65279;&#65279;&#65279;</p>

Explanation / Answer

This is simple vector addition. In order to do this it is best to a) separate your vectors into North/South and East/West components, b) to find the sum of those respective components, and c) to use the Pythagorean theorem to recombine the North/South and East/West components

A) Separate your vectors:
City A-->B
North/South = 0 km
East/West = -200 km [call east positive and west negative]
City B-->C
North/South = 245sin(25) = 103.5 km [call north positive and south negative]
East/West = -245cos(25) = - 222.0 km

B) Add your components:
North/South = 0 + 103.5 = 103.5 km
East/West = -200 -222.0 = -422 km

C) Combine using Pythagorean theorem:
c2= 103.52+(-422)2

c = 434.5 km

Now, you can use trigonometry to find the angle:
tan-1(422/103.5) = 76.2° North of West

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