How is this angle relation true?

Question

Either I'm silly and I'm missing something very simple, or my text book is incorrect. I'm trying to verify a line in the text book which claims that sin(a) = s/r. I can't seem to prove this to myself and its infuriating.

Answer 1

Let $x$ be the angle (in the triangle) that is opposite to side $s$. Then $a+x=180^\circ$.

It is clear that $\sin x=\frac{s}{r}$. But if two angles add up to $180^\circ$, their sines are the same. So $$\sin a=\sin x=\frac{s}{r}.$$

Answer 2

Imagine a circle centered at the triangle vertex at which angle $a$ is located, with radius $r$. The triangle in the picture is a right triangle whose hypotenuse is a radius (its corners touch the edge and the center of the circle).

If we say that the center of the circle is at the origin $(0,0)$ of the plane, then the definition of the sine function for an angle starting at the positive $x$-axis and going counterclockwise (as with your angle $a$) is the ratio of the height of this triangle to its hypotenuse, which is $\frac{s}{r}$.

Answer 3

Let $b$ be adjacent angle of $a$, i.e., $a+b=180$

$\sin(a)=\sin(180-b)=\sin(b)$

From triangle, we have $\sin(b)=s/r$ therefore, $\sin(a)=\sin(b)=s/r$.

Answer 4

we can show the relation via Area of an triangle too.