# Intuition around why Sine of X angle always equals same result.

My understanding so far.

Sine represents a ratio of two sides of an interior angle within a right angle triangle. So given the three lengths of a triangle you can find the sine of any of the 3 interior angles.

Also if you are given the actual angle of an interior angle, you can get the sine using a calculator.

Thus, I deduce from these two statements that on a right triangle any interior angle of a specific number represents a constant raio, whatever the area of the triangle.

I can visualize that in my head to a certain extent, scaling the triangles sides equally, increases the area of the triangle but not the ratio of the sides.

But I'm wondering if I'm missing anything in terms of intuition around this?

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It is hard to tell what is being asked here. The sine is a function, so it outputs the same value every time you give it the same input. Angles in similar triangles have the same trig ratios, as those angles are the same. –  The Chaz 2.0 Dec 14 '11 at 3:59
@TheChaz You're right, my question is vague. One good point that you helped me clarify in your comment, was the fact that sine is a function. Being a function it can't map to multiple results from the same input. I guess I just struggled with the fact that the sine of X degree is always the same. No matter the size of the triangle. I still can't quite grasp the impact that a 90 degree triangle has on maintaining a constant ratio. For example what happens in non-90 degree triangles with sine? –  drc Dec 14 '11 at 4:12
Trig ratios don't have the same visible connection to sides in non-right triangles (there's a word for those - just can't remember it!). The sine of, say, the 45 degree angle in a 35-45-100 triangle would still be 1/sqrt2, but you'd have to draw altitudes to see such ratios. –  The Chaz 2.0 Dec 14 '11 at 4:24
Here's what I suspect is meant. Take two right triangles with the same shape but different sizes; each has a $\theta^\circ$ angle. Look at the ratio of opposite to hypontenuse. Lo and behold, it's the same in both triangles, despite the difference in sizes. So the question would be: why? How is that proved? –  Michael Hardy Dec 14 '11 at 4:26
OBLIQUE triangles! Had to dig out a precal book... :) –  The Chaz 2.0 Dec 14 '11 at 4:26

I think Michael Hardy has rephrased the question well, and I think The Chaz has answered it by referring to similar triangles. If two right triangles both have an angle $\theta$, then they are similar, so the ratio opposite-to-hypotenuse will be the same in both, so the sine depends only on the angle and not on the area.

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The proof assumes that we are in Euclidean geometry - similarity depends on this - so we need the parallel postulate. In non-Euclidean geometry the situation is interestingly different, in that the angle-sum of a triangle depends on its size. It is not clear from the question what kind/sophistication of proof is required. –  Mark Bennet Dec 14 '11 at 8:03
OP says the sine is a ratio of two sides of a right-angle triangle. I think that puts us firmly in Euclidean geometry. But if you want to write up a treatment of trig functions in a non-Euclidean setting, be my guest. –  Gerry Myerson Dec 14 '11 at 9:31
Fair enough. I was just thinking that it might be worth noting what feeds our intuition about these things - then it might be less counterintuitive to imagine other possibilities. –  Mark Bennet Dec 14 '11 at 12:05
Good point.${}$ –  Gerry Myerson Dec 14 '11 at 12:36
What is at stake here is that in euclidean geometry thanks to the parallel axiom we have the notion of similarity: We can scale our figures by an arbitrary factor $\lambda>0$, whereby all incidences stay intact, all lengths are multiplied by $\lambda$, and all angles stay the same. This additional transformation group is not present in other geometries (which apart from the parallel axiom satisfy much the same axioms): You cannot enlarge a spherical triangle "linearly" by a factor $\lambda>0$ whereby all angles stay the same.