Question

Vectors A and B are shown in the figure. Vector C is given by C = B - A. The

magnitude of vector A is 16.0 units, and the magnitude of vector B is 7.00 units. What

is the angle of vector C, measured counterclockwise from the tx-axis?

O A. 73.1°

O B. 22.4°

O C. 16.9°

O D. 292°

O E. 287°

Answer 1

$c=\sqrt{a^2+b^2-2ab\cos C}=16 .19,$

$\frac a{\sin A}=\frac c{\sin C},\implies A=76°,$

$\theta=76°-59°=17°,$

$x=270°+17°=287°~(answer~E).$