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March 13th, 2008, 02:40 PM   #1
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Triangle side lengths



how do u find x?
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March 13th, 2008, 03:39 PM   #2
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Re: Triangle side lengths

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Originally Posted by lax600


how do u find x?
Let c=x

use the law of cosines

c^2=a^2+b^2-2abcosC

In your case a=15, b=16, C=58 deg are known.

Assuming the angles are correct you could also use the law of sines

a/sinA =c/sinC where A=64 deg.

If you don't get the same c both ways, then the data given is incorrect.
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March 13th, 2008, 07:25 PM   #3
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It's pretty obvious that the given measurements are inconsistent. The third angle would be 58, making the triangle isosceles and hence x equal to 16. That's nonsense, since the shortest side can't be opposite the greatest angle.
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