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

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

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

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

$\text{Direction angle}=\tan^{-1}(\dfrac{R_y}{R_x})$

$\text{Direction angle}=\tan^{-1}(\dfrac{13.571}{-4.18733})$

$\text{Direction angle}=-72.8239$

$\text{Direction angle}=107.1476$

$\text{Direction angle}\approx 107.1$

Now,

$|w|=\sqrt{(|u|\cos \theta +|v|\cos \phi)^2+(|u|\sin \theta +|v|\sin \phi)^2}$

$|w|=\sqrt{(10\cos(45)+13\cos 150)^2+(10\sin(45)+13\sin 150)^2}$

$|w|=\sqrt{(10(\dfrac{1}{\sqrt{2}})+13(-\dfrac{\sqrt{3}}{2}))^2+(10(\dfrac{1}{\sqrt{2}})+13(\dfrac{1}{2}))^2}$

$|w|=14.20236$

$|w|\approx 14.20236$

Therefore, the correct option is D.

Answer 2

Answer:

D) |w| = 14.2; θ = 107.1°

Step-by-step explanation:

got it right on edge :)