Question

Let |u| = 10 at an angle of 45° and |v| = 13 at an angle of 150°, and w = u + v. What is the magnitude and direction angle of w?

|w| = 9.4; θ = 72.9°
|w| = 9.4; θ = 107.1°
|w| = 14.2; θ = 72.9°
|w| = 14.2; θ = 107.1°

Answer 1

D) |w| = 14.2; θ = 107.1°

Explanation:

got it right on edge :)

Answer 2

Given:

`|u|=10` at an angle of 45°.

`|v|=13` at an angle of 150°.

`w=u+v`

To find:

Magnitude and direction angle of w.

Solution:

We have,

`w=u+v`

`\theta = 45^\circ, \phi = 150^\circ`

Now,

`u_x=|u|\cos \theta=10\cos 45^\circ=7.071`

`u_y=|u|\sin \theta=u_y=10\sin 45^\circ=7.071`

`v_x=|v|\cos \phi=13\cos 150^\circ=-11.25833`

`v_y=|v|\sin \phi=u_y=13\sin 150^\circ=6.5`

Using these information, we get

`R_x=u_x+v_x=7.071-11.25833=-4.18733`

`R_y=u_y+v_y=7.071+6.5=13.571`

`\text{Direction angle}=\tan^(-1)((R_y)/(R_x))`

`\text{Direction angle}=\tan^(-1)((13.571)/(-4.18733))`

`\text{Direction angle}=-72.8239`

`\text{Direction angle}=107.1476`

`\text{Direction angle}\approx 107.1`

Now,

`|w|=√((|u|\cos \theta +|v|\cos \phi)^2+(|u|\sin \theta +|v|\sin \phi)^2)`

`|w|=√((10\cos(45)+13\cos 150)^2+(10\sin(45)+13\sin 150)^2)`

`|w|=\sqrt{(10((1)/(√(2)))+13(-(√(3))/(2)))^2+(10((1)/(√(2)))+13((1)/(2)))^2}`

`|w|=14.20236`

`|w|\approx 14.20236`

Therefore, the correct option is D.