|
Law of Sines ProofThe Law of Sines is a/(sin A) = b/(sin B) = c/(sin C) = the diameter of the circumscribed circle. Draw an altitude of the triangle (extending side c, if necessary):
From the diagram, it is clear that h=a sin B, and that h= b sin A. (If side c needed to be extended, because angle B is larger than 90 degrees, then note that the sine of B is equal to the sine of the supplement of B, so even though h is to the right of B, it is still true that h=a sin B.) So a sin B = b sin A, a/(sin A) = b/(sin B) = c/(sin C). Now for the second part, that c/(sin C) is equal to the diameter of the circumscribed circle. Consider angle C, which is subtended by chord c of the circle:
Angle C is the same as angle D in the diagram, below, because it is subtended by the same chord. (Or, perhaps angle D is supplementary to angle C, which happens if C is obtuse).
Since triangle ABD is a right triangle, c/(sin D) is the length of AD, the diameter of the circle. Since the measure of C either equals the measure of D or is supplementary to D, their sines are equal, so c/(sin D) = c/(sin C) = the diameter of the circle. This proves the second part. Related pages in this website:
|
|
The webmaster and author of the Math
Help site is Graeme McRae. |