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June 14th, 2010, 07:24 PM   #1
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Sine/Cosine Triangle

Question:
Quote:
Two straight roads diverge at an angle of 70 degrees. Two cars leave the intersection at 7:00 P.M., one traveling at 55 mi/h, and the other at 30 mi/h. How far apart are the cars at 7:15 P.M.
(in miles)? Select the correct answer. Some choices have been to nearest tenth.
I don't understand how to even solve this or if this is even draw out the problem.
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June 14th, 2010, 08:31 PM   #2
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Re: Sine/Cosine Triangle

Quote:
Originally Posted by RMG46
Question:
Quote:
Two straight roads diverge at an angle of 70 degrees. Two cars leave the intersection at 7:00 P.M., one traveling at 55 mi/h, and the other at 30 mi/h. How far apart are the cars at 7:15 P.M.
(in miles)? Select the correct answer. Some choices have been to nearest tenth.
I don't understand how to even solve this or if this is even draw out the problem.
Let's start small.
A car travels in a direction at a given speed, for a certain amount of time. 55mph * .25hours is the distance this car has traveled, so draw a ray on a piece of paper and declare its length to be 55*.25 miles.

Now draw another ray from the same point so that the two rays make an angle of (about) 70 degrees.

/_

Those two characters (the / and the _ ) intersect at an angle close enough to 70, IMO.

This second ray has length 30*.25
label it.

Label the angle between the rays 70degrees.

Use the law of cosines to find the distance between the endpoints of the rays (crap, I should have used "vector" instead of "Ray"... Too hard to edit it from my phone, so maybe a benevolent joker will help me out!)

so yeah. Law of cosines.
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June 14th, 2010, 09:09 PM   #3
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Re: Sine/Cosine Triangle

Oh I see, so SAS is my route to go?
Thanks :]
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June 15th, 2010, 06:37 AM   #4
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Re: Sine/Cosine Triangle

When you know the values of SAS or SSS, use law of cosines. When you know mostly angles, use law of sines.
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June 15th, 2010, 07:22 AM   #5
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What were the choices for the answer?
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June 15th, 2010, 11:27 AM   #6
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Re: Sine/Cosine Triangle

a. 15.9 miles
b. 11.2 miles
c. No correct answer
d. 12.6 miles
e. 13.2 miles (ANSWER)
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June 15th, 2010, 11:44 AM   #7
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I'd prefer answer (c), as the question tells you that the cars leave the intersection, but doesn't say that they don't both use the same road.
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June 15th, 2010, 11:56 AM   #8
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Re:

Quote:
Originally Posted by skipjack
I'd prefer answer (c), as the question tells you that the cars leave the intersection, but doesn't say that they don't both use the same road.
I'd prefer something less cheeky, since there are (possibly) TWO correct answers. And 2 !=0
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June 15th, 2010, 12:36 PM   #9
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If there are two possibilities, choice (e) isn't the correct answer, just a possibility.
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