Answer to Question #288187 in Physics for Nano

Question

Vector a is 6m due north, b is 4 at an angle of 30 degree Celsius above the horizontal what is the resultant vector?

Answer 1

$R_N=6+4\sin30°=8\text{ m},\\ R_E=0+4\cos30°=3.46\text{ m},\\ R=\sqrt{R_N^2+R_E^2}=8.72\text{ m},\\\space\\ \theta=\arctan\dfrac{R_N}{R_E}=66.6°.$