Find the component vectors if the given changes are made in the indicated examples of this section.

In Example 2, change 126.0° to 216.0°.

Data from Example 2

A surveyor’s marker is located 14.4 m from the set pin at a 126.0° standard position angle as shown in Fig. 9.12. Find the x- and y-components of the displacement of the marker from the pin.

Let the pin be at the origin, the initial point of the displacement vector, with the angle in standard position. The displacement vector directed along the x-axis, d , x is related to the displacement vector d, of magnitude d by d_x = d cos 126.0°

Because the displacement vector directed along the y-axis, d , y could also be placed along the vertical dashed line, it is related to the displacement vector d by d_y = dsin126.0 °. Thus, the vectors d_x and d_y have the magnitudes

Therefore, we have resolved the given displacement vector into two components: one directed along the negative x-axis, of magnitude 8.46 m, and the other directed along the positive y-axis, of magnitude 11.6 m. It is also possible to use the reference angle, as long as the proper sign is attached to each component. In this case, the reference angle is 54.0°, and therefore 

dx = 14.4 cos 126.0 = -8.46 m = 14.4 sin 126.0 = 11.6 m