10.2m[48o] + B = 15m[15o]
10.2*cos48+i10.2*sin48+B=15^cos22+i15*sin22.
6.83 + i7.58 + B = 13.91 + i5.62
B=13.91-6.83 + i5.62 - i7.58=7.08-i1.96
B = sqrt(7.08^2 + (-1.96)^2) = 7.35 m.
b. Tan A = Y/X = -1.96/7.08 = -0.27684
A = -15.47o S of E = 344.5o, CCW.
In the sum A+B=C, vector A has a magnitude of 10.2 m and is angled 48.0° counterclockwise from the +x direction, and vector C has a magnitude of 15.0 m and is angled 22.0° counterclockwise from the -x direction. What are (a) the magnitude and (b) the angle (relative to +x) of Vector B? State your angle as a positive number.
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