How do you deal with 120 degree angles?
If ST = 2, QS = 4, and PT = 5, find PQ.
Draw SM parallel to TP ....let M lie on QP
Draw TL perpendicular to SM......let L lie on SM
Draw QN perpendicular to TP....let QN intersect SM at O and TP at N
In triangle TLS ....since SM is parallel to TP....then angle TLS = 90
And since TL is perpendicular to SM...angle LTP = 90 so angle LTS = 120 - 90 = 30
And angle TSL = 60
So triangle TSL = 30-60-90.... TS = 2 , LS = 1 and TL =sqrt (3)
And in triangle QOS, angle QOS = 90
And angle OSQ = 120 -angle LST = 120 -60 = 60
So triangle SOQ also is 30- 60- 90
And QS = 4 so OS= 2 and QO = 2sqrt 3
And OS - LS = 2 - 1 = 1 = LO = TN
And NP = PT - TN = 5 - 1 = 4
So triangle QNP is a right triangle with leg NP = 4 and leg QN = TL + QO = 3 sqrt 3 = sqrt (27)
So PQ = sqrt [ 4^2 + (sqrt (27))^2 ] = sqrt [ 16 + 27 ] = sqrt (43)