An airplane flies 200 km due west from city A to city B and then 290 km in the direction of 27.5° north of wes

Question

An airplane flies 200 km due west from city A to city B and then 290 km in the direction of 27.5° north of west from city B to city C.

a) In straight-line distance, how far is city C from city A?

b) Relative to city A, in what direction is city C?
° north of west

Answer 1

Such problem can be solved systematically with a coordinate system. Let us have x-axis pointing to the east and y-axis pointing to the north. Let us also call the vector from A to B as AB, vector from B to C as BC, and the resulting vector from A to C as AC. As you know, the unit vector along the x-axis is i and along the y-axis is j.
AB = -200i
BC = -290cos27.5°i + 290sin27.5°j
Thus AC = AB + BC =
= (-290cos27.5° - 200)i + 290sin27.5°j
= ai + bj
So the distance |AC| = sqrt(a^2 + b^2)
and the direction of AC: arc tan (b/a)

Answer 2

a bigger triangle could be formed via extending the west line to a factor under the trip spot giving a discern with vert. part n and base of (200+w) Then: n = 225* sin(35.) and w = 225* cos(35.5) the gap A - C is then Sqrt(n^2 + (w+200)^2))