why soh cah toa is right?

Question

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 ,

https://qph.is.quoracdn.net/main-qimg-83442b1d2c8b9f5962f427dc2d46fea7?convert_to_webp=true

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

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

Answer 1

$\sin \angle A$ 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 $\triangle APQ\sim\triangle ABC$:

$$\begin{align*} \frac{PQ}{BC} &= \frac{AP}{AB}\\ \frac{\sin\angle A}{BC} &= \frac{1}{AB}\\ \sin \angle A &= \frac{BC}{AB}\\ BC &= AB \sin \angle A \end{align*}$$

Answer 2

I have taught soh-cah-toa in the context of right triangles where the sides can be any length. In that context (geometry), the sine is the ratio of the side opposite the angle over the hypotenuse, whatever the lengths of the sides may be.

In trigonometry, the angle is measured on the unit circle, where the radius of the circle (and also the hypotenuse) is 1. Therefore, sine is the length of the opposite side when the hypotenuse is 1. The sine of a given angle is always the same due to similar triangles. The way we taught how sine is defined changes based on context, but the numeric value stays the same.