In the picture, the unknown you are looking for (the distance between the two cities) is unlabeled, but is opposite of the angle of 2.1^o2.1o that is shown at the satellite. For convenience sake I will label the angle of 2.1^o2.1o as /_C∠C, and the unknown distance between the cities as cc. Furthermore, I will arbitrarily label the "top" distance of 370km as aa and the "bottom" distance of 350km as bb.
In this triangle problem, we have 2 sides (aa and bb) and the angle between those sides (/_C∠C), and we are solving for the opposite side (cc), which means the Law of Cosines is best suited for this problem:
c^2 = a^2 + b^2 -2abcos Cc2=a2+b2−2abcosC
Filling in the knowns as listed above, we have to solve for cc, which should be a simple matter so long as care is taken to calculate all values properly. (This is particularly true when using a calculator when some questions may be in radians and others in degrees!)
c^2 = a^2 + b^2 -2abcos Cc2=a2+b2−2abcosC
c^2 = (370)^2 + (350)^2 - 2(370)(350)(cos 2.1) c2=(370)2+(350)2−2(370)(350)(cos2.1)
c^2 ~~ 136900 + 122500 - 259000(0.999) c2≈136900+122500−259000(0.999)
c^2 ~~ 259400 - 258826.054 c2≈259400−258826.054
c^2 ~~ 573.946 c2≈573.946
c ~~ sqrt(573.946) ~~ 24.0"km" c≈√573.946≈24.0km
