Answer:
584.3 m at [tex]135.7^{\circ}[/tex] above the positive x-axis
Explanation:
We have to find the magnitude and direction of the resultant vector. In order to do that, we have to resolve each vector along the x- and y- direction first.
Resolving the vector L:
[tex]L_x = L cos \theta_L = 303 \cdot cos 205^{\circ} =-274.6m\\L_y = L sin \theta_L = 303 \cdot sin 205^{\circ} = -128.1 m[/tex]
Resolving the vector M:
[tex]M_x = M cos \theta_M = 555 \cdot cos 105^{\circ} =-143.6 m\\M_y = M sin \theta_M = 555 \cdot sin 105^{\circ} = 536.1 m[/tex]
Now we can find the components of the resultant vector by adding the components of each vector along each direction:
[tex]R_x = L_x + M_x = -274.6 +(-143.6)=-418.2 m\\R_y = L_y + M_y = -128.1 +536.1 = 408 m[/tex]
So the magnitude of the resultant vector is
[tex]R=\sqrt{R_x^2 + R_y^2}=\sqrt{(-418.2)^2+(408)^2}=584.3 m[/tex]
While the direction is given by:
[tex]\theta = tan^{-1} \frac{R_y}{R_x}=tan^{-1} \frac{408}{418.2}=44.3^{\circ}[/tex]
But since [tex]R_x[/tex] is negative and [tex]R_y[/tex] is positive, it means that this angle is measured as angle above the negative x-axis; so the direction of the vector is actually
[tex]\theta = 180^{\circ} - 44.3^{\circ} = 135.7^{\circ}[/tex]
