# why soh cah toa is right?

i am confused by the sine of an angle, (it might appear evident for some of you but please i am not an expert ). sine of an angle is says to be the half of the magnitude of the chord of 2 time the angle.geometrically, it is just the opposite edge of a right triangle. so if one do not associate a circle , one angle will have infinite sine. then why the formula soh cah toa is still applicable since it imply the use of two circles(each of the similar right triangles)? considering this representation ,

depend on the circle , sin Â = BC or sin Â = B'C'. how can sin Â be (B'C')/(AB')?

thank in advance, i will be happy to reformulate the question if there is an unclear point

• The circle in that chord definition is a unit circle, with radius $1$. – peterwhy Apr 27 '16 at 21:23
• thank you, may i ask what it changes? – Tony Tam Apr 27 '16 at 21:36
• That means there is only one such circle, not infinite ones with arbitrary radii. – peterwhy Apr 27 '16 at 21:42
• then the way to think about it is (refereeing to the previous picture) AB=AB'=1? – Tony Tam Apr 27 '16 at 21:53
• If the hypotenuse is $1$, then yes, the opposite side is the sine of that angle. But if not, then the opposite side is no longer that sine, but scaled according to similar triangle laws. – peterwhy Apr 27 '16 at 21:55 $\sin \angle A$ is only the (signed) length of the opposite side for a hypotenuse $1$. To obtain the length of general $BC$, consider corresponding sides of the similar triangles $\triangle APQ\sim\triangle ABC$:
\begin{align*} \frac{PQ}{BC} &= \frac{AP}{AB}\\ \frac{\sin\angle A}{BC} &= \frac{1}{AB}\\ \sin \angle A &= \frac{BC}{AB}\\ BC &= AB \sin \angle A \end{align*}