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v and a are angles and s and r are the sides of the right triangle 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.

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    There are several definitions of the sine of an obtuse angle, and your definition has to be your starting point. (If you *have* no definition, then it’s impossible to understand the statement.)2012-08-29

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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 $\sin a=\sin x=\frac{s}{r}.$

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    @QEntanglement: There is another way of putting it which is visually useful: The sine of $90^\circ+t$ is the same as the sine of $90^\circ -t$, that is, the sine function is *symmetric* about $90^\circ$ (or in radians, $\frac{\pi}{2}$ radians).2012-08-29
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we can show the relation via Area of an triangle too.

enter image description here

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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 $\frac{s}{r}$.

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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, $\sin(a)=\sin(b)=s/r$.