The vector with modulus ( length ) r, and in the direction inclined at
#theta^o# to x-axis in the anticlockwise ( + ) sense, has
#(r, theta)
#as the polar #(|..|, angle)# form,
but component-wise, it is
#r < cos theta, sin theta > = < x, y > #.
Here,
#vec A= (15, 0^o) = 15 <cos 0^o,5 sin 0^0 >=15<1, 0> # and
#vec B= (30, 60^o) = < 30 cos 60^o, 30 sin 60^o >#
# =<15, 15sqrt3>=15 <1, sqrt3>#.
Now, with component-wise addition and subtraction,
#vecA + vec B = 15<1, 0>+15<1, sqrt3>=15<2, sqrt3># and
#vec B - vec A =15<1, sqrt3> -15<1, 0> = 15<0, sqrt3>#
The polar #(|..|, angle)# forms are
#vec A + vec B =( 15sqrt7, 40.89^o)#, nearly, and
#vec B - vec A= (15sqrt3, 90^o)#.
