We are asked to determine the sum of the two vectors F1 and F2. To do that we will use the following triangle:
Therefore, substituting the magnitudes of the vectors the triangle is:
Now, to determine the magnitude of the resultant force "R" we will use the cosine law:
[tex]c^2=a^2+b^2-2ab\cos\theta[/tex]
Where:
[tex]\begin{gathered} c=\text{ opposite side to the 120\degree angle} \\ a,b=\text{ other sides of the triangle} \end{gathered}[/tex]
Now, we substitute the values:
[tex]R^2=5^2+5.9^2-2(5)(5.9)\cos(121)[/tex]
Solving the operations we get:
[tex]R^2=90.25[/tex]
Now, we take the square root to both sides:
[tex]\begin{gathered} R=\sqrt{90.25} \\ R=9.5 \end{gathered}[/tex]
Therefore, the magnitude of the sum of vectors is 9.5
Now, we determine the angle:
We need to determine angle "y". To do that we will determine the angle "x" using the sine law:
[tex]\frac{\sin A}{a}=\frac{\sin B}{b}[/tex]
Where:
[tex]\begin{gathered} a=\text{ side opposite to angle A} \\ b=\text{ ide oppoite to angle B} \end{gathered}[/tex]
Now, we plug in the known values:
[tex]\frac{\sin x}{5}=\frac{\sin121}{9.5}[/tex]
Now, we multiply both sides by 5:
[tex]\sin x=5\frac{\sin121}{9.5}[/tex]
Solving the operations:
[tex]\sin x=0.45[/tex]
Now, we take the inverse function of the sine:
[tex]x=\sin^{-1}(0.45)[/tex]
Solving the operations:
[tex]x=26.83[/tex]
Now, we have that since F1 is a vertical force then the sum of angles "x" and "y" must add up to 90:
[tex]x+y=90[/tex]
Now, we substitute the value of angle "x":
[tex]26.83+y=90[/tex]
Now, we subtract 26.83 from both sides:
[tex]\begin{gathered} y=90-26.83 \\ y=63.17 \end{gathered}[/tex]
Therefore, the angle of the sum of vectors is 63.17°.