Draw the -8km vector from the origin on the neg. x-axis.
Draw the 6km vector from the origin 45o
west of north(135o).
Du = 6km @ 135o.
Du = 6*cos135 + i6*sin135
Du = -4.24 + i4.24 = Dist. from Oshawa
to Uxbridge.
Da = -8 km=Dist. from Oshawa to Ajax.
D = Du - Da
D = (-4.24+i4.24) - (-8)
D = 3.76 + i4.24
D^2 = (3.76)^2 + (4.24)^2 = 32.12
D = 5.67 km.
Draw a diagram to solve this problem: Ajax is 8 km due west of Oshawa. Uxbridge is 6 km NW of Oshawa. How far is it from Ajax to Uxbridge? Explain whether you have enough information to solve this problem?
2 answers
post it.