Recent content by ejantz

ooo okay, then what about this: 13.1769 (cos^2 θ1 + sin^2 θ1) = 29.04 cos θ1 - 13.144 13.1769 = 29.04 cos θ1 - 13.144 cos θ1 = (13.1769 + 13.144) / 29.04 θ1 = cos-1 (0.906) θ1 = 25.0 θ1 + θ2 = 90 θ2 = 90 - (25.0) θ2 = -25 woww that was such a struggle, I am sorry, I think I may have it...

ah no sorry i completely forgot about the other half of the equation! ignore the previous post, check this: (4 - v'1 cosθ1)^2 + (v'1 sinθ1)^2 = 1.69^2 (cos^2 θ2 + sin^2 θ2) (4 − v′1 cosθ1)^2 + ((3.63)^2 X sin^2 θ1) = 2.8561 (4 − v′1 cosθ1)^2 --> v'1^2 cos^2 θ1 − 8v'1 cos θ1 +16...

ohhh right, okay: (4 − v′1 cosθ1)^2 = 2.8561 v'1^2 cos^2 θ1 − 8v'1 cos θ1 +16 = 2.8561 (3.63)^2 cos^2 θ1 - 8(3.63)cos(θ) + 16 = 2.8561 13.1769 cos^2 θ1 - 29.04 cos θ1 + 13.1439 apply the quadratic equation for: 13.1769 x^2 - 29.04 x + 13.1439 = 1.57 and 0.63 θ1 = cos-1 0.63...

okay perfect, yes. so next step: (4 - v'1 cosθ1)^2 + (v'1 sinθ1)^2 = 1.69^2 (cos^2 θ2 + sin^2 θ2) this second part would be: = 2.8561 X 1 = 2.8561 but the first half now I am still struggling.. should i substitute in the v'1 (3.63) value? (4^2 - 3.63^2 X cos^2 θ1) + (3.63^2 X sin^2 θ1)...

I know this question has been asked before but I really need some help finishing this final piece of the puzzle.. I have attached an image of my work to show how far I have gotten towards the solution but my unfamiliarity with trigonometry equations has gotten my stuck. Please assist! thank you