How to solve this VECTORS?

If this is a trianle ABC, angleCAB=40; angleABC=127.
By using the cos law: c^2=a^2+b^-2abcosC
Distance You are looking for= "b" =the length of the diagonal opposite to the angleABC (127 deg).
->b^2=2^2+4^2-2(2)(4)cos127
After knowing b, you have the data for all three diagonals plus 2 angles. You have to look for angleBCA to find the direction.
->2^2=4^2+b(u found it)^2-2(4)(b)cosC

whilst including vectors, it fairly is appropriate to apply real effective angles and to apply X Y diagrams. east=0 stages north=ninety stages west=one hundred eighty stages south=270 stages to function vectors, you may convert polar to rectangular coordinates, then upload them. 10 stages N of E=10 stages 30 stages S of E=(-30) stages from 0 stages or 330 stages (real). So: one hundred twenty(cos 10)i+one hundred twenty(sin 10)j a hundred and fifty(cos 330)i+a hundred and fifty(sin 330)j (via utilising real angles, you do no longer could concern approximately + and - signs and indicators, by using fact the calculator will do it for you, once you put in the real attitude.) After including the two vectors above, use Pythagorean to locate the cost and the arc tan functionality to locate the attitude. (Remembering tan=sin/cos