My Math Forum Detailed section of all triangles which have unexplained properties

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 January 11th, 2019, 06:56 PM #11 Global Moderator   Joined: Dec 2006 Posts: 21,127 Thanks: 2336 The inequalities $a+b>c,\, a+c>b,\, b+c>a$ are known as the triangle inequality. Your third and fourth identities can be written as $(c\times\sin B)^2+(c\times\cos B)^2=c^2$ and $(c\times\sin A)^2+(c\times\cos A)^2=c^2$ so that they hold for any value of $c$, not just when $c$ is 1. Hence my change to this effect in your previous list. As $c\times\sin B = b\times\sin C$, $(c\times\sin B)^2+(b\times\cos C)^2=b^2$ is equivalent to $(b\times\sin C)^2+(b\times\cos C)^2=b^2$, which can be simplified to $\sin^2\! C + \cos^2\! C = 1$. I think all the identities you listed can be found on Wikipedia. Thanks from topsquark
 January 13th, 2019, 10:52 AM #12 Newbie   Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0 Correction for the fourth line. $(b\times\cos A)^2+(a\times\cos B)^2=c^2$
 January 13th, 2019, 02:15 PM #13 Newbie   Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0 For side c so the base is not only 1the equations are: $(c\times\sin B)^2+(c\times\cos B)^2=c^2$ $(c\times\sin A)^2+(c\times\cos A)^2=c^2$
January 13th, 2019, 02:17 PM   #14
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Quote:
 Originally Posted by skipjack The inequalities $a+b>c,\, a+c>b,\, b+c>a$ are known as the triangle inequality. Your third and fourth identities can be written as $(c\times\sin B)^2+(c\times\cos B)^2=c^2$ and $(c\times\sin A)^2+(c\times\cos A)^2=c^2$ so that they hold for any value of $c$, not just when $c$ is 1. Hence my change to this effect in your previous list. As $c\times\sin B = b\times\sin C$, $(c\times\sin B)^2+(b\times\cos C)^2=b^2$ is equivalent to $(b\times\sin C)^2+(b\times\cos C)^2=b^2$, which can be simplified to $\sin^2\! C + \cos^2\! C = 1$. I think all the identities you listed can be found on Wikipedia.
Thanks

 January 13th, 2019, 07:50 PM #15 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Regarde dans le dictionnaire Larousse
January 13th, 2019, 10:52 PM   #16
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Quote:
 Originally Posted by Larrousse Correction for the fourth line. $(b\times\cos A)^2+(a\times\cos B)^2=c^2$
That's incorrect. I'm not sure what you intended.

January 14th, 2019, 01:52 AM   #17
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 Originally Posted by skipjack That's incorrect. I'm not sure what you intended.
It's an error, disregard it.

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