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

  • Thread starter FawkesCa
  • Start date
You are not a retard, this problem was just poorly worded and had a small error in it.In summary, the conversation discusses a math problem involving finding the width of a stream given certain measurements and angles. The participants discuss their attempts at solving the problem and eventually come to the conclusion that there may be an error in the problem itself. The conversation also highlights the importance of being confident in one's own work and seeking help when needed.
  • #1
FawkesCa
43
0
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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  • #2


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.
 
  • #3


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.
 
  • #4


Likewise, 400.739 ft., by

[tex]w = \frac{d \; \sin(\beta) \; \sin(\alpha)}{\sin(180^{\circ}-\alpha-\beta)}[/tex]
 
  • #5


[tex]\sin(180^{\circ}-\alpha-\beta)=\sin(\alpha+\beta)[/tex] :smile:
 
  • #6


Ooh, so it is. Goodbye pi! So long, minuses!
 
  • #7


And with a little re-arrangement, bring on the cotangents that I used! :biggrin:
 
  • #8


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.
 
  • #9


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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