Given:
Vector A = 63.5 m
Vector B = 101 m at an angle of 57.0 degrees.
Let's find the magnitude of the sum of these two vectors.
To find the magnitude of the sum of the two vector, we have:
Vector A.
X-component and y-component of vector A.
[tex]\begin{gathered} A_x=63.5cos90=0\text{m} \\ \\ A_y=63.5sin90=63.5\text{ m} \end{gathered}[/tex]
Vector B.
x- and y-component of vector B:
[tex]\begin{gathered} B_x=101cos57=55\text{ m} \\ \\ B_y=101sin57=84.7\text{ m} \end{gathered}[/tex]
Now, to find the magnitude of the sum of the vectors, we have:
[tex]\begin{gathered} \sqrt{(A_x+B_x)^2+(A_y+B_y)^2} \\ \\ \sqrt{(0+55)^2+(63.5+84.7)^2} \end{gathered}[/tex]
Solving further:
[tex]\begin{gathered} \sqrt{3025+(148.2)^2} \\ \\ \sqrt{3025+21963.24} \\ \\ \sqrt{24988.24} \\ \\ =158.08\text{ m} \end{gathered}[/tex]
Therefore, the magnitude of the sum of these two vectors is 158.08 m.
ANSWER:
