In the sum →A+→B=→C, vector →A has a magnitude of 12.0 m and is angled 40.0° counterclockwise from the + x direction, and vector →C has a magnitude of 15.0 m and is angled 20.0° counterclockwise from the - x direction. What are (a) the magnitude and (b) the angle (relative to + x) of →B?

Explanation: Ok, first caracterize the two vectors that we know.

A = ax + ay = (12*cos (40°) * i + 12*sin (40°) * j) m

now, see that C is angled 20° from - x, - x is angled 180° counterclockwise from + x, so C is angled 200° counterclockwise from + x

C = cx + cy = (15*cos (200°) * i + 15*sin (200°) * j) m

where i and j refers to the versors associated to te x axis and the y axis respectively.

in a sum of vectors, we must decompose in components, so: ax + bx = cx and ay + by = cy. From this two equations we can obtain B.

bx = (15*cos (200°) - 12*cos (40°)) m = - 23.288 m

by = (15*sin (200°) - 12*sin (40°)) m = - 12.843 m

Now with te value of both components of B, we proceed to see his magnitude an angle relative to + x.

Lets call a to the angle between - x and B, from trigonometry we know that tg (a) = by/bx, that means a = arctg (12.843/23.288) = 28.8°

So the total angle will be 180° + 28.8° = 208.8°.

For the magnitude of B, lets call it B', we can use the angle that we just obtained.

bx = B'*cos (208.8°) so B' = (-23.288 m) / cos (208.8°) = 26.58 m.

So the magnitude of B is 26.58 m.