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 October 5th, 2018, 07:00 AM #1 Senior Member   Joined: Nov 2011 Posts: 248 Thanks: 3 law of sines Why the author use the law of sines? What is goal to use it? In the link: https://www.maa.org/external_archive...n/Polygon.html
 October 5th, 2018, 07:43 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,303 Thanks: 1974 That possibly wasn't intended.
October 5th, 2018, 07:54 AM   #3
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Quote:
 Originally Posted by skipjack That possibly wasn't intended.
Excuse me ?!

October 5th, 2018, 02:11 PM   #4
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Quote:
 Originally Posted by shaharhada Why the author use the law of sines? What is goal to use it? In the link: https://www.maa.org/external_archive...n/Polygon.html
The law of sines gives the length of the side opposite the angle, needed to compute the area of the triangle whose other two sides are radii of length one.

Last edited by skipjack; October 6th, 2018 at 03:24 AM.

 October 6th, 2018, 04:19 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,303 Thanks: 1974 It brings in the sine of another angle, which isn't in the desired area formula. Area.PNG By the above diagram, $A(P_nOP_{\small1}) = \frac121\cdot1\cdot\sin(\alpha) = \frac12\sin(\alpha)$.
 October 6th, 2018, 02:10 PM #6 Global Moderator   Joined: May 2007 Posts: 6,684 Thanks: 659 Area formula (radius=1). Split triangle in half (right triangles) altitude $=\cos(\frac{\alpha}{2})$, base$=\sin(\frac{\alpha}{2})$ Area of original triangle $=\cos(\frac{\alpha}{2})\sin(\frac{\alpha}{2}) = \frac{1}{2}\sin(\alpha)$. Last edited by skipjack; October 6th, 2018 at 04:11 PM.
 October 6th, 2018, 05:01 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,303 Thanks: 1974 The half angle identity used above is most easily proved by use of the $\frac12bc\sin(A)$ formula for a triangle's area.

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