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Michael and Zach are hiking in the mountains. They left Michael’s car in the parking lot and walked northwest for 12.4 km to a campsite. Then they turned due south and walked another 7.0 km to a glacier lake. The weather was taking a turn for the worse, so they decided to plot a course directly back to the parking lot. Zach remembered, from the map in the parking lot, that the angle between the path to the campsite and the path to the glacier lake measures about 30°. What compass direction should they follow to return directly to the parking lot?

Sagot :

Answer:

  about 101°

Step-by-step explanation:

The bearing back to the parking lot is the reverse of the bearing from the parking lot to the lake. That bearing can be found two ways. One of them uses the bearing and distance of each leg of the hike. The other uses what Zach remembers about the map.

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sum of vectors

The location of the lake from the parking lot can be found by adding the vectors to the campsite from the lot and to the lake from the campsite. That sum will be ...

  12.4∠315° +7.0∠180° ≈ 8.94∠281.4°

A suitable calculator can be used to find this sum, as shown in the second attachment. (Angles are displayed by the calculator in the range (-180°, 180°].)

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The direction of the parking lot from the lake will be the opposite direction. It can be found by subtracting 180° from the angle:

  bearing to parking lot = 281.4° -180° = 101.4°

Michael and Zack should follow a compass direction of about 101°.

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map memory

Northwest is a compass bearing of 315°. Points south of that bearing will have a smaller angle. Zach remembers the direction of the lake to be about 30° less than that angle, so at 315° -30° = 285°.

The bearing back to the parking lot is the opposite direction, so ...

  285° -180° = 105°

Based on Zack's memory, they should follow a compass direction of about 105°.

They will come within about 560 meters of the parking lot using this heading.

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Additional comments

When doing these vector calculations by hand, a couple of different methods can be used. One is to convert the vectors to component form, add the components, then convert back to polar form. The other is to make use of the Law of Cosines and the Law of Sines to solve the relevant triangles.

Here, the lake-lot distance is the third side of a triangle with two sides and the included angle known. The Law of Cosines applied to this situation is ...

  c² = a² +b² -2ab·cos(C) . . . . . generic Law of Cosines equation

  c² = 12.4² +7.0² -2(12.4)(7.0)cos(45°) ≈ 80.0063

  c ≈ √80.0063 ≈ 8.9446

Using the Law of Sines, we find the obtuse angle of the triangle to be ...

  sin(L)/12.4 = sin(C)/8.9446 . . . . . L and C are vertex angles in the diagram

  L = 180° -arcsin(12.4/8.9446×sin(45°)) ≈ 101.4°

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If translation to rectangular coordinates is used, this can be done using the usual translation ...

  (x, y) = (r·cos(θ), r·sin(θ)) . . . . rectangular coordinates for r∠θ

The value of θ used can be the bearing angle (measured clockwise from north), or it can be the corresponding Cartesian plane angle (measured counterclockwise from +x). As long as the angles are consistently measured, the math works either way.

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We have assumed that "northwest" is a bearing of 315°. If the compass rose is divided into 32 segments, the name "northwest" will be given to any bearing within 5° of this value. This makes our detailed vector calculations be somewhat approximate, and means that Zach's memory of the required bearing may be sufficiently accurate after all. Or not.

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