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^o# that is shown at the satellite. For convenience sake I will label the angle of #2.1^o# as #/_C#, and the unknown distance between the cities as #c#. Furthermore, I will arbitrarily label the "top" distance of 370km as #a# and the "bottom" distance of 350km as #b#.
In this triangle problem, we have 2 sides (#a# and #b#) and the angle between those sides (#/_C#), and we are solving for the opposite side (#c#), which means the Law of Cosines is best suited for this problem:
#c^2 = a^2 + b^2 -2abcos C#
Filling in the knowns as listed above, we have to solve for #c#, 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 C#
#c^2 = (370)^2 + (350)^2 - 2(370)(350)(cos 2.1) #
# c^2 ~~ 136900 + 122500 - 259000(0.999) #
# c^2 ~~ 259400 - 258826.054 #
# c^2 ~~ 573.946 #
# c ~~ sqrt(573.946) ~~ 24.0"km" #
