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allochka39001 [22]
3 years ago
12

An airplane has its auto pilot set to fly in a direction of 40 degrees at a speed of 320 mph. The wind is blowing the airplane t

owards a direction of 130 degrees at a speed of 20 mph. If the auto pilot does not account for the wind that is pushing against the plane find the exact distance and direction of the plane after 1 hour of flying.

Mathematics
1 answer:
Effectus [21]3 years ago
8 0

Answer:

distance: 320.624 miles

direction: 43.576°

Step-by-step explanation:

The speed and direction can be found by adding the given vectors.

... 320∠40° + 20∠130°

... = (320·cos(40°), 320·sin(40°)) + (20·cos(130°), 20·sin(130°))

... = (245.134, 205.692) +(-12.856, 15.321) = (232.278, 221.013)

The magnitude of the vector with these components is found using the Pythagorean theorem. The direction is found using the arctangent function.

... = √(232.278² +221.013²)∠arctan(221.013/232.278)

... = 320.624∠43.576°

_____

A suitable vector or graphing calculator can do this easily. In the screenshot of a TI-84 app below, the variable D has the value π/180. The display mode is set to degrees.

_____

<em>Comment on coordinate systems</em>

Navigation directions are generally measured clockwise from North. Angles in the usual x-y coordinate plane are measured counterclockwise from +x (effectively, East). You can consider the geometry of the navigation coordinate system to be a reflection across the line y=x of the geometry of the usual x-y coordinate system.

Reflection does not change lengths or angles within a given geometry. Hence, we can use all the usual tools of vector calculation using navigation coordinates, without bothering to translate them back and forth to x-y coordinates.

_____

Problems like this generally can be worked using the Law of Cosines and the Law of Sines, too. It generally helps to draw a diagram so you can find the values of the angles betwee the various vectors more easily.

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Gre4nikov [31]

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Answer:

  1. Translate P to E; rotate ∆PQR about E until Q is coincident with F; reflect ∆PQR across EF
  2. Reflect ∆PQR across line PR; translate R to G; rotate ∆PQR about G until P is coincident with E

Step-by-step explanation:

The orientations of the triangles are opposite, so a reflection is involved. The various segments are not at right angles to each other, so a rotation other than some multiple of 90° is involved. A translation is needed in order to align the vertices on top of one another.

The rotation is more easily defined if one of the ∆PQR vertices is already on top of its corresponding ∆EFG vertex, so that translation should precede the rotation. The reflection can come anywhere in the sequence.

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<em>Additional comment</em>

The mapping can be done in two transformations: translate a ∆PQR vertex to its corresponding ∆EFG point; reflect across the line that bisects the angle made at that vertex by corresponding sides.

3 0
2 years ago
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<span>m∠cab = 2x
m∠​acb = x + 30
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Hello I am happy to help you with this problem. So the formula to solve for a triangle is A = bxh divided by 2.
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I hope that helps you. If you have any questions please feel free to ask me.
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