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Dahasolnce [82]
2 years ago
15

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?

Mathematics
2 answers:
mixer [17]2 years ago
4 0

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.

Elis [28]2 years ago
4 0

Answer:

D) |w| = 14.2; θ = 107.1°

Step-by-step explanation:

got it right on edge :)

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