The Student Room Group
Reply 1
call vectors A and B, and let A be horizontal. then A = Xi, B = Xcosyi + Xsinyj, where y is the angle between them.

12 = sqrt((X+Xcosy)^2 + Xsiny^2)
144 = X^2 + 2X^2 cosy + x^2 cos^2 y + X^2 sin^2 y
144 = X^2(1+ 2cosy + cos^2 y + 1 - cos^2 y) = X^2(2+2cosy) = 2X^2(1+cosy)

bloody hell

21 = sqrt((1.7X+Xcosy)^2 + Xsiny^2)
441 = 2.89X^2 + 3.4X^2cosy + x^2 cos^2 y + X^2 sin^2 y = X^2(2.89 + 3.4cosy + cos^2 y + 1 - cos^2 y) = X^2(3.89 + 3.4cosy)

X^2 = 441/(3.89 + 3.4cosy) = 72/(1+cosy)
441 + 441cosy = 280.08 + 244.8cosy
y = acos(-160.92/196.2) = about 145 degrees

then plug in to find x.

but what horrible numbers. maybe i'm wrong.
Reply 2
ummm . .. isnt there a less complicated way to solve this problem !!!
Reply 3
yes, of course.

draw the two triangles, see you get another with lengths 21, 12 and 0.7X. work with that instead.