Question

two pairs of hikers leave the same camp heading in opposite divections. Each travels 2 miles, then changes direction and travels 1.2 miles. The first pair starts due east and then turns 50° toward north. The second pair starts due west and then turns 40° toward south. Which pair is farther from camp? Explain your reasoning.

Answer 1

As a result, the first pair of hikers is around 0.15 miles further away from camp than the second pair.

Explanation:

Let's designate "x" for the distance travelled directly east or west and "y" for the distance directly north or south. Then, using the method below, determine how far each pair of hikers is from camp:

Distance is equal to sqrt(2xy*cos(angle) - xy + yy).

where "angle" is the angle at which the hikers veer away from their original path.

The first group of hikers includes:

Distance is equal to sqrt(((2cos(50°) + 1.2sin(50°))).

(1.2cos(50°) + 2 sin(50°))

(1.2sin(50°) + 2cos(50°)) = 2 - 2*

(1.25*cos(50°) + 1.2*cos(50°)*cos(130°))

approximately 2.53 miles.

The second group of hikers includes:

Distance is equal to sqrt(((2cos(40°) + 1.2sin(40°))).

-2sin(40°) + 1.2cos(40°) = 2

(1.2sin(40°) + 2cos(40°)) = 2 - 2*

(-2sin(40°) plus 1.2*cos(40°)*cos(140°))

about 2.38 miles.

As a result, the first pair of hikers is around 0.15 miles further away from camp than the second pair.