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If I have an angle on a right triangle that is $\pi$ degrees, that would mean that the length of the opposite side would have to equal $0$, because:

$\sin(\pi) = \frac {\text{opposite}} {\text{hypotenuse}}$

$0 = \frac {\text{opposite}} {\text{hypotenuse}}$

$0 \times \text{hypotenuse} = \text{opposite}$

$0 = \text{opposite}$

When I set an angle of a right triangle to pi breaks all the rules I thought I knew. What is the explanation of all this

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    $\begingroup$ A triangle with one angle equal to $\pi$ will be a line (at least in Euclidean geometry). $\endgroup$
    – Dante
    Nov 2 '16 at 20:08
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    $\begingroup$ $\sin(\pi)=0$ when you are measuring the angle in radians, not degrees. $\endgroup$
    – Joffan
    Nov 2 '16 at 20:08
  • $\begingroup$ $\sin\pi=0$ in radians, not degrees. It's the same thing as saying $\sin180^\circ=0$. You can't have a triangle with an angle of $180^\circ$, just as you can't have a triangle with a side of length $0$. $\endgroup$
    – Jam
    Nov 2 '16 at 20:08
  • $\begingroup$ $\pi$ degrees? Do you mean radians? If so, think about how many degrees are in $\pi$ radians. $\endgroup$ Nov 2 '16 at 20:08
  • $\begingroup$ A right angle has measure of $\frac \pi2$ radians, and in degrees, $90^\circ$. $\endgroup$
    – amWhy
    Nov 2 '16 at 20:15
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When you learned the sine function using the definition $$ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}, $$ you learned about a function whose domain is just the values $\theta$ such that $0 < \theta < \frac\pi2$.

If you are OK with defining certain degenerate triangles a certain way, you can extend this to $0 \leq x \leq \frac\pi2$. But you cannot construct a right triangle with angles outside that domain, at least not in ordinary Euclidean geometry.

The sine function that mathematicians actually use, however, is defined over all the real numbers. There are various ways to do this; one that I like is to draw a unit circle around the origin of an $x,y$ Cartesian plane, and then, starting at $x=1, y=0$, travel a distance $\theta$ counterclockwise around the circle to reach a point $(x_\theta,y_\theta)$. Then $\sin\theta = y_\theta$.

For $0 < \theta < \frac\pi2$, the triangle $(0,0), (x_\theta,0), (x_\theta,y_\theta)$ is a right triangle such that the angle at $(0,0)$ is $\theta$, the hypotenuse of the triangle is $1$, and the side opposite $(0,0)$ is $y_\theta$. Hence for those angles, $$ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y_\theta}{1} = y_\theta, $$ which agrees with the definition you learned. That is, the mathematician's sine function is just an extension of the function you learned to real numbers less than $0$ and greater than $\frac\pi2$.

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A right angle has measure of $\frac \pi 2$ radians, and in degrees, it is $90^\circ$.

The two other angles of a right triangle must add up to $\frac \pi2$ radians, or $90^\circ$, and this is because the sum of the measures of the angles of any/every triangle equals $\pi$ radians, or equivalently, $180^\circ$.

A triangle with one angle of $\pi$ radians, or equivalently $180^\circ$ is what we call a "degenerate triangle", a "triangle" that degenerates to a straight line.

Usually, when given the measure of an angle a multiple of $"\pi"$, we assume the angle is measured in radians. If $x$ is given in radians, then $\frac{x\cdot 180^\circ}{\pi}$is the measure in degrees.

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  • $\begingroup$ Note that $\sin(\pi) = 0$ only when we're working with $\pi$ radians. $\sin(\pi^\circ) = 0.05480366517\neq 0$. Bottom line, if you are going to use "degrees", do so consistently, and make sure your calculator is set to "degrees". Alternatively, if you're going to use radians, do so consistently, and set your calculator to "radian" mode. $\endgroup$
    – amWhy
    Nov 2 '16 at 20:41
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    $\begingroup$ I think my confustion came from thinking that sin( pi degrees) = 0, when, instead, it is sin(pi radians) = 0 $\endgroup$ Nov 3 '16 at 14:09

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