If you have the diagram, you can see that ∡ is 16°
So use the law of cosines.
GE^2 = 160^2 + 200^2 - 2*160*200 cos16°
The diagram shows the position of three airports . A and G. G is 200 kilometres from A E is 160 kilometres from A From G the bearing of A is 052 degrees From A the bearing of E is 216 degrees How far apart are airports G and E?
3 answers
I got the answer to be 158km but im not sure how to explain it can someone help me
All angles are measured CW from +y-axis.
AG = 200km[52o], GA = 200km[52o+180o].
AE = 160km[216o].
GE = GA + AE = 200[232] + 160[216].
GE = (200*sin232+160*sin216) + (200*cos232+160*cos216)I
-252 - 253i.
GE = sqrt(252^2 + 253^2) =
AG = 200km[52o], GA = 200km[52o+180o].
AE = 160km[216o].
GE = GA + AE = 200[232] + 160[216].
GE = (200*sin232+160*sin216) + (200*cos232+160*cos216)I
-252 - 253i.
GE = sqrt(252^2 + 253^2) =