Hello, Barthayn!

$\displaystyle \text{6. Each of three brothers has tied a rope to a buoy floating on a lake.}$

$\displaystyle \,A\text{ pulls towards the east with a force of 40 N.}$

$\displaystyle \,B\text{ pulls towards the southwest with a force of 30 N.}$

$\displaystyle \text{What force should }C\text{ exert to hold the buoy in equilibrium?}$

$\displaystyle \text{The answer is: }\, 28.3\,N\text{ at }131^o.$

Code:

C
o
*
*
*
*
o * * * * * o A
* O 40
30 *
*
o
B

We have: .$\displaystyle \begin{Bmatrix}\overrightarrow{OA} &=& \langle 40,0\rangle \\

\overrightarrow{OB} &=& \langle\text{-}15\sqrt{2},\text{-}15\sqrt{2}\rangle \\ \overrightarrow{OC} &=& \langle x,y\rangle \end{Bmatrix}$

We want: .$\displaystyle \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} \:=\:\overrightarrow{0} $

. . $\displaystyle \langle 40,0\rangle + \langle\text{-}15\sqrt{2},\text{-}15\sqrt{2}\rangle + \langle x,y\rangle \;=\;\langle 0,0,0\rangle$

. . $\displaystyle \langle x + 40 - 15\sqrt{2},\;y - 15\sqrt{2}\rangle \;=\;\langle 0,0\rangle$

$\displaystyle \text{Then: }\;\begin{Bmatrix} x+40 - 15\sqrt{2}\:=\:0 & \Rightarrow &x \:=\:15\sqrt{2}-40 \\ y - 15\sqrt{2} \:=\:0 & \Rightarrow & y \:=\:15\sqrt{2} \end{Bmatrix}$

$\displaystyle |\overrightarrow{OC}| \;=\;\sqrt{x^2+y^2} \;=\;\sqrt{\left(15\sqrt{2}-40\right)^2 + \left(15\sqrt{2}\right)^2}$

. . . $\displaystyle =\;28.33626167 \;\approx\;\boxed{28.3\text{ N}}$

$\displaystyle \tan\theta \:=\:\dfrac{y}{x} \;=\;\dfrac{15\sqrt{2}}{15\sqrt{2} - 40} \;=\;-1.129154902$

. . $\displaystyle \theta \;=\;131.5286818 \;\approx\;\boxed{131.5^o}$