The first vector #vec(V_1)# has a magnitude of 11.7 m and is angled 35.8° counterclockwise from the +x direction.
so #vec(V_1)=11.7cos35.8hati+11.7sin35.8hatj#
#=>vec(V_1)=9.5hati+6.8hatj#
The 2ndt vector #vec(V_2)# has a magnitude of16.5 m and is angled 13.0° counterclockwise from the -x direction.
so #vec(V_2)=16.5cos193hati+16.5sin193hatj#
#=>vec(V_2)=-16.0hati-3.7hatj#
So resultant vector
#=>vec(V)=vec(V_1)+vec(V_2)=-7.5hati+3.1hatj#
(a) so magnitude of resultant vector
#=>abs(vec(V))=abs(-7.5hati+3.1hatj)=8.1m#
(b) And the resultant makes an angle
#tan^-1(-3.1/7.5)=157.5^@#counter clockwise with +x axis
