The route followed by a hiker consists of three displacement vectors A with arrow, B with arrow, and C with arrow. Vector A with arrow is along a measured trail and is 1550 m in a direction 25.0° north of east. Vector B with arrow is not along a measured trail, but the hiker uses a compass and knows that the direction is 41.0° east of south. Similarly, the direction of vector vector C is 20.0° north of west. The hiker ends up back where she started, so the resultant displacement is zero, or A with arrow + B with arrow + C with arrow = 0. Find the magnitudes of vector B with arrow and vector C with arrow.

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lcmendozaf lcmendozaf · Answer 1 · 2019-09-15T00:49:41+02:00

Answer:

[tex]D_{B}=1173.98m\\D_{C}=675.29m[/tex]

Explanation:

If we express all of the cordinates in their rectangular form we get:

A = (1404.77 , 655.06) m

[tex]B = A + ( -D_{B} *sin(41) , -D_{B} * cos(41) )[/tex]

[tex]C = A + B + ( -D_{C} *cos(20) , D_{C} * sin(20) )[/tex]

Since we need C to be (0,0) we stablish that:

[tex]C = (0,0) = A + B + ( -D_{C} *cos(20) , D_{C} * sin(20) )[/tex]

That way we make an equation system from both X and Y coordinates:

[tex]A_{x} + B_{x} + C_{x} = 0[/tex]

[tex]A_{y} + B_{y} + C_{y} = 0[/tex]

Replacing values:

[tex]1404.77 - D_{B}*sin(41) - D_{C}*cos(20) = 0[/tex]

[tex]655.06 - D_{B}*cos(41) + D_{C}*sin(20) = 0[/tex]

With this system we can solve for both Db and Dc and get the answers to the question:

[tex]D_{B}=1173.98m[/tex]

[tex]D_{C}=675.29m[/tex]