The route followed by a hiker consists of three displacement vectors A with arro

Question

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 27.0° 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 24.0° 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.

Explanation / Answer

All x-component vectors and y-component vectors need to add up to ZERO.

X: Acos(27°) - Bsin(41°) - Ccos(24°) = 0 [1]

Y: Asin(27°) - Bcos(41°) + Csin(24°) = 0 [2]

A = 1550 m

Now we have 2 equations and 2 unknowns, so we solve for B and C:

Equation [1] becomes

C = [Acos(27°) - Bsin(41°)] / cos(24°)

Substitute C into equation [2]

Asin(27°) - Bcos(41°) + ([Acos(27°) - Bsin(41°)] / cos(24°)) sin(24°) = 0

Asin(27°) + Acos(27°) tan(24°) = Bcos(41°) + Bsin(41°) tan(24°)

A(sin(27°)+cos(27°)tan(24°)) = B(cos(41°)+sin(41°)tan(24°))

Therefore,

B = A(sin(27°)+cos(27°)tan(24°)) / (cos(41°)+sin(41°)tan(24°))

= 672.6 m

C = 1028.7 m

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