How Far Should the Rescue Crew Fly to Reach the Plane?

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Homework Help Overview

The problem involves a plane's flight path consisting of two legs at specified angles and distances, culminating in an emergency landing. The task is to determine the direct distance a rescue crew must fly to reach the plane.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the correct interpretation of angles and distances in the problem, with some suggesting the use of the law of cosines to find the direct distance. Others question the geometric setup and the validity of applying the Pythagorean theorem.

Discussion Status

There is an ongoing exploration of the correct angles and the application of geometric principles. Some participants have provided guidance on using the law of cosines, while others are clarifying assumptions about the triangle formed by the flight path.

Contextual Notes

Participants are addressing potential misinterpretations of the angles described in the problem, which may affect the calculations. The discussion is framed within the constraints of a homework assignment, emphasizing the need for careful consideration of the problem's setup.

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Emergency Landing. A plane leaves the airport in Galisto and flies a distance 155 km at an angle 68.0 Degrees east of north and then changes direction to fly a distance 270 km at an angle 48.0 degrees south of east, after which it makes an immediate emergency landing in a pasture.

1.) When the airport sends out a rescue crew, how far should this crew fly to go directly to this plane?

how do i find that? i included a picture that i THINK is correct.
 

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The direction of the first leg of the trip is incorrect in your picture. It says 68 degrees east of north in the problem, but your picture shows 68 degrees north of east.

Once you've cleared that up, the law of cosines should help you find the length in question (first use your knowledge of geometry to determine the angle).
 
am i trying to find the third leg? if so can't i do 155^2+270^2=C^2
C=311 .32
 
I'm afraid not. That would only work if this were a right triangle. You should find that the angle opposite the third leg is 110 degrees. The law of cosines goes like this:

[tex]c^2 = a^2 + b^2 - 2ab\cos\theta[/tex]

where θ is the angle opposite side c.
 
ah thanks a lot. one more question...
In which direction? (Express your answer as an angle measured south of east).

how would i find out which direction it's going?
 

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