Respuesta :
Answer:
magnitude = 10.8
angle = 118 degree
Explanation:
Magnitude of vector B is 5.35 m and angle is 60 degree
so we have
[tex]\vec B = 5.35 cos60\hat i + 5.35 sin60 \hat j[/tex]
[tex]\vec B = 2.675 \hat i + 4.63 \hat j[/tex]
Now let the magnitude of vector A is "x" and it makes and angle theta
so we have
[tex]\vec A = x cos\theta \hat i + x sin\theta \hat j[/tex]
now for vector C we have
[tex]\vec C = x cos(\theta + 25) \hat i + x sin(\theta + 25)[/tex]
now we know that
[tex]\vec A. \vec B = 2.675 x cos\theta + 4.63 x sin\theta[/tex]
[tex]30.6 = x(2.675 cos\theta + 4.63 sin\theta)[/tex]
also we know that
[tex]\vec B. \vec C = 2.675 x cos(\theta + 25) + 4.63 x sin(\theta + 25)[/tex]
[tex]36.9 = x(2.675 cos(\theta + 25) + 4.63 sin(\theta + 25))[/tex]
now we have
[tex]\theta = 118 degree[/tex]
[tex]x = 10.8 m[/tex]
The magnitude and direction of vector A are approximately 7.011 meters and 24.670°.
Let be [tex]A(x,y) = \|A\|\cdot (\cos \theta_{A},\sin \theta_{A})[/tex], [tex]B(x,y) = \|B\|\cdot (\cos \theta_{B},\sin \theta_{B})[/tex] and [tex]C(x,y) = \|A\|\cdot (\cos (\theta_{A}+25^{\circ}),\sin (\theta_{A}+25^{\circ}))[/tex], we obtain the magnitude and direction of vector A by definition of dot product:
[tex]\|A\|\cdot \|B\|\cdot (\cos \theta_{A}\cdot \cos \theta_{B} + \sin \theta_{A}\cdot \sin \theta_{B}) = 30.6\,m^{2}[/tex] (1)
[tex]\|B\|\cdot \|A\|\cdot (\cos \theta_{B}\cdot \cos (\theta_{A}+25^{\circ})+\sin\theta_{B}\cdot \sin (\theta_{A}+25^{\circ})) = 36.9\,m^{2}[/tex] (2)
By equalizing (1) and (2), we get the following expression:
[tex]\frac{\cos \theta_{A}\cdot \cos \theta_{B}+\sin \theta_{A}\cdot \sin \theta_{B}}{30.6} = \frac{\cos \theta_{B}\cdot \cos (\theta_{A}+25^{\circ})+\sin \theta_{B}\cdot \sin (\theta_{A}+25^{\circ})}{36.9}[/tex]
If we know that [tex]\theta_{B} = 60^{\circ}[/tex], then we have the following expression:
[tex]\frac{0.5\cdot \cos \theta_{A}+0.866\cdot \sin \theta_{A}}{30.6} = \frac{0.5\cdot \cos (\theta_{A}+25^{\circ})+0.866\cdot \sin (\theta_{A}+25^{\circ})}{36.9}[/tex]
[tex]36.9\cdot (0.5\cdot \cos \theta_{A}+0.866\cdot \sin \theta_{A}) = 30.6\cdot [0.5\cdot \cos (\theta_{A}+25^{\circ})+0.866\cdot \sin (\theta_{A}+25^{\circ})][/tex]
[tex]18.45\cdot \cos \theta_{A}+31.955\cdot \sin \theta_{A} = 15.3\cdot \cos (\theta_{A}+25^{\circ})+26.5\cdot \sin (\theta_{A}+25^{\circ})[/tex]
[tex]18.45\cdot \cos \theta_{A} + 31.955\cdot \sin \theta_{A} = 15.3\cdot (\cos \theta_{A}\cdot \cos 25^{\circ}-\sin \theta_{A}\cdot \sin 25^{\circ})+26.5\cdot (\sin \theta_{A}\cdot \cos 25^{\circ} + \sin 25^{\circ}\cdot \cos \theta_{A})[/tex]
[tex]18.45\cdot \cos \theta_{A} + 31.955\cdot \sin \theta_{A} = 13.867\cdot \cos \theta_{A}-6.466\cdot \sin \theta_{A}+24.017\cdot \sin \theta_{A}+11.199\cdot \cos \theta_{A}[/tex]
[tex]18.45\cdot \cos \theta_{A}+31.955\cdot \sin \theta_{A} = 25.066\cdot \cos \theta_{A}+17.551\cdot \sin \theta_{A}[/tex]
[tex]14.404\cdot \sin \theta_{A} = 6.616\cdot \cos \theta_{A}[/tex]
[tex]\tan \theta_{A} = 0.459[/tex]
[tex]\theta_{A} \approx 24.670^{\circ}[/tex]
By (1) we get the magnitude of vector A: ([tex]\|B\| = 5.35[/tex], [tex]\theta_{A} \approx 24.670^{\circ}[/tex], [tex]\theta_{B} = 60^{\circ}[/tex])
[tex]\|A\| = \frac{30.6}{\|B\|\cdot (\cos \theta_{A}\cdot \cos \theta_{B}+\sin \theta_{A}\cdot \sin \theta_{B})}[/tex]
[tex]\|A\| = \frac{30.6}{5.35\cdot (\cos 24.670^{\circ}\cdot \cos 60^{\circ}+\sin 24.670^{\circ}\cdot \sin 60^{\circ})}[/tex]
[tex]\|A\| \approx 7.011\,m[/tex]
The magnitude and direction of vector A are approximately 7.011 meters and 24.670°.
We kindly invite to check this question on dot product: https://brainly.com/question/16537974
