Question 818134
Here's one way to solve this...<ol><li>On a graph, label the origin as "O".</li><li>Plot the point (15, 0) and label it V. This makes OV an image of vector v.</li><li>In component form, v = 15i + 0j.</li><li>Draw a segment from O at an angle of 120 degrees from OV. This segment should extend into the second quadrant. Label the end of this segment as U and label its length as 12. This make OU an image of vector u.</li><li>Draw a vertical segment down from U to the negative part of the x-axis. Label this point A.</li><li>Draw a segment from A to O. This completes the triangle OUA. And because UA is vertical and OA is horizontal, UA and OA form a right angle. This makes triangle OUA a right triangle.</li><li>Since angle UOV is 120 degrees, angle UOA is 60 degrees (because these two angles form a linear pair which are always supplementary).</li><li>Since angle UOA is 60 degrees and triangle UOA is a right triangle, triangle UOA is a 30/60/90 right triangle.</li><li>Since triangle UOA is a 30/60/90 right triangle we can use the pattern for this type of triangle to find the remaining sides. With a hypotenuse of 12, the side opposite the 30 degree angle, OA, will be half as much: 6. Label the length of OA as 6.</li><li>And the side opposite 60 degrees is {{{sqrt(3)}}} times the side opposite 30: {{{sqrt(3)*6}}} or {{{6sqrt(3)}}}. Label the length of UA as {{{6sqrt(3)}}}</li><li>Using UA and OA we can now express vector u in component form:
{{{u = -6i + 6sqrt(3)j}}} (Note: The i component is negative because it goes to the left from the origin.)</li><li>We can now add vectors u and v using the components:
{{{u + v = (-6i + 6sqrt(3)j) + (15i + 0j) = 9i + 6sqrt(3)j}}}</li><li>We can now find the magnitude of u+v:
{{{abs(u + v) = sqrt((9)^2 + (6sqrt(3))^2) =  sqrt(81+36*3) = sqrt(81+108) = sqrt(189)}}} which is approximately 13.7477</li><li>We can also now find the direction:
{{{theta = tan^-1(6sqrt(3)/9) = tan^-1(2sqrt(3)/3)}}} which is approximately 49.1066 degrees.</li></ol>