Question 1149454
A=10N
B=12N
theta=60 degrees
C={{{sqrt(A^2+B^2+2ABcos(theta))}}}
Plugging in the values we get, C=19.0787N
Angle of the resultant={{{ arctan(Bsin(theta)/(A+Bcos(theta))) }}}
Plugging in the values we get, angle=33.0044 degrees