If $\to B$ $vec B$ is added to $\to C = 4.8 ˆ i + 4.3 ˆ j$ $vec C=4.8hati+4.3hatj$ , the result is a vector in the positive direction of the y axis, with a magnitude equal to that of $\to C$ $vecC$ . What is the magnitude of $\to B ?$ $vecB?$

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If $\to B$ $vec B$ is added to $\to C = 4.8 ˆ i + 4.3 ˆ j$ $vec C=4.8hati+4.3hatj$ , the result is a vector in the positive direction of the y axis, with a magnitude equal to that of $\to C$ $vecC$ . What is the magnitude of $\to B ?$ $vecB?$

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Answer 1 · 2017-09-13T05:41:49

The magnitude of $\to B$ $vecB$ is $= 5.25$ $=5.25$

Explanation:

The magnitude of $\to C$ $vecC$ is

$∣ ∣ ∣ ∣ ∣ ∣ \to C ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ 4.8 ˆ i + 4.3 ˆ j ∣ ∣ ∣ ∣ = \sqrt{{4.8}^{2} + {4.3}^{2}} = 6.44$ $||vecC||=||4.8hati+4.3hatj||=sqrt(4.8^2+4.3^2)=6.44$

The angle is $θ = arctan (\frac{4.3}{4.8}) = {41.86}^{\circ}$ $theta=arctan(4.3/4.8)=41.86^@$

Applying the cosine rule to the isoceles triangle,

$B^{2} = C^{2} + C^{2} - 2 C^{2} cos (90 - 41.86)$ $B^2=C^2+C^2-2C^2cos(90-41.86)$

$B^{2} = 2 C^{2} (1 - cos (90 - 41.86)) = 27.6$ $B^2=2C^2(1-cos(90-41.86))=27.6$

$B = \sqrt{27.6} = 5.25$ $B=sqrt(27.6)=5.25$

The magnitude of $\to B$ $vecB$ is $= 5.25$ $=5.25$

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If $\to B$ $vec B$ is added to $\to C = 4.8 ˆ i + 4.3 ˆ j$ $vec C=4.8hati+4.3hatj$ , the result is a vector in the positive direction of the y axis, with a magnitude equal to that of $\to C$ $vecC$ . What is the magnitude of $\to B ?$ $vecB?$

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Explanation:

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If →Bvec B is added to →C=4.8ˆi+4.3ˆjvec C=4.8hati+4.3hatj, the result is a vector in the positive direction of the y axis, with a magnitude equal to that of →CvecC. What is the magnitude of →B?vecB?

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Explanation:

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If $\to B$ $vec B$ is added to $\to C = 4.8 ˆ i + 4.3 ˆ j$ $vec C=4.8hati+4.3hatj$ , the result is a vector in the positive direction of the y axis, with a magnitude equal to that of $\to C$ $vecC$ . What is the magnitude of $\to B ?$ $vecB?$