IbTrojan Posted May 12, 2015 Report Share Posted May 12, 2015 I have another question. It's attached to this post, I got (a) and (b) but I'm not entirely sure how to get ©... If someone could please help in any way, I would appreciate it. Thanks! Reply Link to post Share on other sites More sharing options...
IbTrojan Posted May 13, 2015 Author Report Share Posted May 13, 2015 Anyone? Please? Paper 2 is tomorrow...and it'd be nice to know how to figure this out before writing the exam. Reply Link to post Share on other sites More sharing options...
Emmi Posted May 13, 2015 Report Share Posted May 13, 2015 Cosine rule: a^2 = b^2 + c^2 - 2*b*c*cosASine rule: a/sinA = b/sinB = c/sinC A:AC^2 = AB^2 + BC^2 - 2*AB*BC*cos(AC)= 4^2 + 5^2 - 2*5*4*cosx= 16 + 25 - 40cosx= 41 - 40cosxThus AC = sqrt(41 - 40cosx)B:AC/sin(AC) = BC/sin(BC) AC/sinx = 4/sin(30)AC/sinx = 4/0.5 = 8AC = 8sinxC:8sinx = sqrt(41 - 40cosx)64sin^2x = 41 - 40cosx64sin^2x + 40cosx = 4164(1 - cos^2x) + 40cosx - 41 = 0-64cos^2x + 40cosx + 23 = 064cos^2x - 40cosx - 23 = 0Use the quadratic formula:Cosx = (40 +/- sqrt(40^2 - 4*64*-23))/2*64= (40 +/- 86.533)/128= 0.9885408 or - 0.3635408X = arccos(0.9885408) or arccos(-0.3635402) which ends up giving us x = 8.68 degrees or x = 111.37 degrees. We want the larger angle here so x = 111.37.Now just use either of the two relations we found for AC:AC = sqrt(41 - 40cos(111.37)) = 7.45 cmAC = 8sin(111.37) = 7.45 cm 1 Reply Link to post Share on other sites More sharing options...
IbTrojan Posted May 13, 2015 Author Report Share Posted May 13, 2015 Thank you Emmi! Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.