Originally Posted by

**rebghb** Ok it's quite simple, it goes like:

$\displaystyle F_1+F_2=(45i+0j) + (60\cos\theta\,i+60\sin\theta\,j)=(45+60\cos\theta )\,i + 60\sin\theta\,j$ which is equal to the resultant **vector. **

Then it follows that $\displaystyle (45+60\cos\theta)^2 \ + \ (60\sin\theta)^2=90^2$ and thus by expanding the squares we get **$\displaystyle 45^2+60^2\cos^2\theta+(45)(60)(2)\cos\theta+60^2\, \sin^2\theta=90^2$ and by manipulation we get that $\displaystyle \cos\theta=\frac{-11}{24} \Leftrightarrow \theta=117.27 \mbox{degrees}$.**