My Math Forum Triangle side lengths

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 March 13th, 2008, 03:40 PM #1 Newbie   Joined: Mar 2008 Posts: 6 Thanks: 0 Triangle side lengths how do u find x?
March 13th, 2008, 04:39 PM   #2
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Re: Triangle side lengths

Quote:
 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.

 March 13th, 2008, 08:25 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,983 Thanks: 1853 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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