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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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It’s pretty much just the definition of $\sin a$. What seems unreasonable about it? – Brian M. Scott Aug 29 '12 at 20:47
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.) – Lubin Aug 29 '12 at 23:58
up vote 5 down vote accepted

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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Thanks!! I did not know that simple fact that, "If two angles add up to 180, then their sines are the same." – QEntanglement Aug 29 '12 at 21:22
@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). – André Nicolas Aug 29 '12 at 21:25

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

enter image description here

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


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

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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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