Answer to Question #91492 in Physics for Karanwadalkar Karanwadalkar

Question

Vector A of magnitude 8 unit at 30°north of West and another vector B of magnitude 4unit at 60° south of East find the magnitude and direction of vector A+B?

Answer 1

$${\vec C}={\vec A}+{\vec B}$$

$$C_x=A_x+B_x=-8\cos(30^{\circ})+4\cos(60^{\circ})=2-4\sqrt{3}$$

$$C_y=A_y+B_y=8\sin(30^{\circ})-4\sin(60^{\circ})=4-2\sqrt{3}$$

The magnitude of A+B

$$C=\sqrt{C_x^2+C_y^2}=\sqrt{(2-4\sqrt{3})^2+(4-2\sqrt{3})^2}=4.96$$

The direction of A+B

$$\theta=\tan^{-1}\frac{-C_y}{C_x}=\tan^{-1}\frac{-4+2\sqrt{3}}{2-4\sqrt{3}}$$

$$=6.25^{\circ}\quad \rm{North\;of\; West}$$