sines law proof
Let ABC a triangle. Let T a point in the circumcircle such that BT is a diameter.
So ∠A=∠CAB is equal to ∠CTB (they subtend the same arc). Since △CBT is a right triangle, from the definition of sine we get
sin∠CTB=BCBT=a2R. |
On the other hand ∠CAB=∠CTB implies their sines are the same and so
sin∠CAB=a2R |
and therefore
asinA=2R. |
Drawing diameters passing by C and A will let us prove in a similar way the relations
bsinB=2R |
and we conclude that
Q.E.D.
Title | sines law proof |
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Canonical name | SinesLawProof |
Date of creation | 2013-03-22 11:57:36 |
Last modified on | 2013-03-22 11:57:36 |
Owner | drini (3) |
Last modified by | drini (3) |
Numerical id | 9 |
Author | drini (3) |
Entry type | Proof |
Classification | msc 51-00 |
Related topic | CosinesLaw |
Related topic | Triangle |
Related topic | SinesLaw |