Calculating Stream Width from AB to Point C | Law of Sines Application

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Homework Help Overview

The problem involves calculating the width of a stream using the Law of Sines, given a line segment AB of 652 feet and angles A and B. The original poster expresses frustration over a discrepancy between their calculated width and the answer provided in a reference book.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss their calculations and results, noting that they arrive at a similar width of approximately 400 feet. There is a focus on the interpretation of the angles A and B, with some questioning whether these angles are correctly defined in relation to the triangle formed.

Discussion Status

Multiple participants have confirmed the same calculated width, indicating a shared understanding of the problem setup. There is ongoing exploration of the definitions of angles and the application of trigonometric identities. Some guidance is offered regarding the need to verify assumptions and seek clarification from instructors.

Contextual Notes

Participants mention potential typos in textbooks and express concerns about the clarity of the problem statement. The original poster's repeated calculations suggest a strong confidence in their method, despite conflicting answers from the reference material.

FawkesCa
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for God's sake...I NEED HELP!

Homework Statement


AB is a line 652 feet long on one bank of a stream, and C is a point on the opposite bank. A = 53° 18', and B = 48° 36'. Find the width of the stream from C to AB.


Homework Equations



law of sines

The Attempt at a Solution


ive proven all variables are correct (A,B,C and a, b, c). i come up with a height of 400.74ft. the answer book says its 345.45ft.
ive done this 6 times and come up with the same answer. am i a retard or should i just not go to Clark University (its their website)
 
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I'm also getting the same answer as you, if you let the two angles be [tex]\alpha[/tex],[tex]\beta[/tex] and the distance across the bank be d, then the width of the river is given by

[tex]w=\frac{d}{cot(\alpha)+cot(\beta)}[/tex]

and plugging in your values gives approx 400ft.
 


I get the same answer as the OP as well. But there is some ambiguity as to what the angles are. The "A = some angle" and "B = some angle" statements really don't make sense, since A and B are just corners of the triangle. I assumed that "A" actually meant angle CAB, and that "B" actually meant angle CBA.
 


Likewise, 400.739 ft., by

[tex]w = \frac{d \; \sin(\beta) \; \sin(\alpha)}{\sin(180^{\circ}-\alpha-\beta)}[/tex]
 


[tex]\sin(180^{\circ}-\alpha-\beta)=\sin(\alpha+\beta)[/tex] :smile:
 


Ooh, so it is. Goodbye pi! So long, minuses!
 


And with a little re-arrangement, bring on the cotangents that I used! :biggrin:
 


thank you, thank you, THANK YOU! i guess clark university is not the best place to try and teach yourself trigonometry. all of you help me SOOOOOOO much.
 


FawkesCa said:
i guess clark university is not the best place to try and teach yourself trigonometry.

Not necessarily. Pretty much every book is going to have a few typos in it, you just need to be prepared. If you are getting a different answer and are sure of yourself, then don't be so quick to disbelieve your own work. Leave it and ask your professor or here online.
 

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