Question

Displacement vectors A, B, and C add up to a total of zero. Vector A has a magnitude of 1550 m and a direction of 24.7 north of east. Vector B has a direction of 41.0° east of south, and vector C has a direction of 33.9 north of west. Find the magnitudes (in, m) of vector B and vector C. magnitude of B magnitude of C

Answer 1

Answer:

|B|=5078.62m   and   |C|=5710.83m

Explanation:

In order to solve this problem, we must first make a drawing of what the vectors will look like. (See picture uploaded)

on picture a) we can see the vectors mentioned on the question, each with its corresponding direction relative to the coordinate axis, or the compass.

As you may see, a triangle is formed by the three vectors. This is because they add up to a total of zero, which means they will all go back to the origin.

On picture b) we can see how the angles are used to find the inner angles of the given triangles. All I did there was find complementary angles and locate alternate angles, notice they are all found with respect to the coordinate system.

On picture c) we can see the final triangle with the corresponding angles a, b and c.

Once we got the angles of the triangle we can solve it by using the law of sines:

$\frac{A}{sin a}=\frac{B}{sin b}=\frac{C}{sin c}$

so we can use that to find the magnitude of B and C:

$\frac{B}{sin b}=\frac{A}{sin a}$

Which solves to:

$B=\frac{Asin b}{sin a}$

$B=\frac{(1550m)sin (58.6^{o})}{sin (15.1^{o})}$

so |B|=5078.62m

And the magnitude of C is found similarly:

$\frac{C}{sin c}=\frac{A}{sin a}$

Which solves to:

$C=\frac{Asin c}{sin a}$

$C=\frac{(1550m)sin (106.3^{o})}{sin (15.1^{o})}$

so |C|=5710.83m