Vector c has a magnitude of 22.2 m and points in the −y‑ direction. vectors a and b both have positive y‑ components, and make angles of α=41.9° and β=28.2° with the positive and negative x- axis, respectively. if the vector sum a+b+c=0 , what are the magnitudes of a and b?

The magnitude of A is 17.46 m and B is 1.50 m

Step-by-step explanation:

If the vector sum A+B+C =0, then the sum of the projection of the vector in axes x- is zero and the sum of the projection of the vector in the axes y- is also zero.

Ax+Bx+Cx = 0

Ay+By+Cy = 0

|Ax| = cos 41.9 * |A|

|Ay| = sin 41.9 * |A|

|Bx| = cos 28.2 * |B|

|By| = sin 28.2 * |B|

|Cx| = 0

|Cy| = 22.2

Ax+Bx+Cx = 0

|Ax|-|Bx|+0 =0

the vector Ax is in the positive direction of the x- axes and Bx in the negative direction and C do not have a component in the x- axes

cos 41.9 * |A| - cos 28.2 * |B| = 0 (I)

Ay+By+Cy = 0

|Ay|+|By|-|Cy|=0

the vector Ay and By are the positive direction of the y- axes and Cy in the negative direction

sin 41.9 * |A| + sin 28.2 * |B| - 22 =0 (II)

Now we have a system of 2 (I and II) equations and 2 variables (|A| and |B|)

cos 41.9 * |A| - cos 28.2 * |B| = 0

sin 41.9 * |A| + sin 28.2 * |B| = 22

cos 41.9 * |A| = cos 28.2 * |B|

|A| = cos 28.2 * |B| / cos 41.9

sin 41.9 * |A| + sin 28.2 * |B| = 22

sin 41.9 *  cos 28.2 * |B| / cos 41.9 + sin 28.2 * |B| = 22

tg 41.9 * cos 28.2 * |B| + sin 28.2 * |B| = 22

(tg 41.9 * cos 28.2 + sin 28.2) * |B| = 22

|B| = 22 / (tg 41.9 * cos 28.2 + sin 28.2)

|B| = 17.46

|A| = 1.50

yes

step-by-step explanation:

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