Given:
Vector A = 63.5 m at 90 degrees
Vector B = 101 m at 57.0 degrees
Let's find the direction of the sum of these two vectors.
To find the direction, let's first find the x- and y-components of the vectors.
• Vector A:
[tex]\begin{gathered} A_x=65.3cos90=0\text{ m} \\ A_y=63.5sin90=63.5\text{ m} \end{gathered}[/tex]
• Vector B:
[tex]\begin{gathered} B_x=101cos57.0=55\text{ m} \\ B_y=101sin57=84.7\text{ m} \end{gathered}[/tex]
For the sum of components, we have:
x = Ax + Bx = 0 + 55 = 55 m
y = Ay + By = 63.5 + 84.7 = 148.2 m
Now, to find the direction of the sum, we have:
[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
Plug in the values and solve for θ.
We have:
[tex]\begin{gathered} \theta=tan^{-1}(\frac{148.2}{55}) \\ \\ \theta=69.6^o \end{gathered}[/tex]
Therefore, the direction of the sum of the vectors is 69.6 degrees.
ANSWER:
69.6 degrees.
