By placing the vector as the hypotenuses of a right triangle we can see the distance can be broken up into two components (vertical and horizontal). These can be found with trig:
[tex]\begin{gathered} vertical=80\cos (65)=33.809\rightarrow down \\ \text{horizontal}=80\sin (65)=72.504\rightarrow right \end{gathered}[/tex]
Doing the same for the other force we get:
[tex]\begin{gathered} vertical=100\cos (62)=46.947\rightarrow up \\ \text{horizontal}=100\sin (62)=88.294\rightarrow right \end{gathered}[/tex]
Adding up all vertical and horizontal we get:
[tex]\begin{gathered} vertical\colon33.809+46.947=80.756 \\ \text{horizontal: 72.504+88.294=160.798} \end{gathered}[/tex]
The magnitude is the hypotenuse. Therefore:
[tex]\sqrt[]{80.756^2+160.798^2}=179.94[/tex]
Answer: 179.94 pounds
