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Sum of angles of a triangle is 180 degress. So while studying trigonometric ratios , I got surprised at cos0 and cos 180 values. Although cosine is ratio of adjecent and hypotenuse ,in case of cos0 with value 1 or cos180 with value 0, i doubt that how can it be called as triangle where one angle is 0 or 180 degree?

It would be just be just a straight line rather than called as triangle,isn't it? Also I want to know what could be the minimum and maxmimum value of angle in the triangle ?

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    If you do not accept degenerate triangles, then there is no minimum and no maximum – Hagen von Eitzen Jul 03 '18 at 05:34
  • @Hagen von Eitzen suppose you are expaining trigonometry topic to a person . Let us take triangle ABC and take cosine of angle C . In one case angle is 0 degree in other case angle is 180 degree. So person will ask directly ask that " is it really called as triangle?". Then what should youreply? – user3559780 Jul 03 '18 at 05:41
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    @user35597890 I would reply that these are called degenerate triangles. These may be frowned upon as not being real triangles, but they may occur as limiting cases, for example if we want to define trigonometric functions of $0^\circ$ and $180^\circ$ - and we have not even disciussed how to define for negative angles or those beyond $180^\circ$! – Hagen von Eitzen Jul 03 '18 at 05:46
  • This answer includes a discussion that might help with understanding why and how we define sine and cosine at $0^\circ$ and $90^\circ$ (and beyond the First Quadrant). – Blue Jul 03 '18 at 06:58

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What you described at the first part is what we call degeneracy, some people accept degenerate triangle as triangle, but if you don't we have no minimum (nor maximum) angle:

Give me a triangle with an angle $\alpha$ I can create a triangle with an angle $\alpha/2$ and a(different) triangle with angle $(\alpha+180)/2$(degrees), the first triangle shows that $\alpha$ is not minimal angle, and second shows that $\alpha$ is not maximal angle.

ℋolo
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The usual "SOH CAH TOA" definition of trigonometry is only useful in non-degenerate right triangles (usually first discussed in Geometry in the U.S). And in this case, you are right. It's hard to imagine a right triangle with $0^\circ$ or $180^\circ$, or even worse, negative angles or angles greater than $180^\circ$. In order to approach these angles, you need new definition of trigonometry (although still related to "SOH CAH TOA"). You need "Unit Circle Approach". Let $P(x,y)$ be a terminal point on a unit circle centered at origin where we moved a distance $\vert\theta\vert$ along its arc, starting at the point $(1,0)$. We define: $\sin\theta=y$, $\cos\theta=x$, $\tan\theta=\frac{y}{x}$.

If $\theta$ is exactly $180^\circ$, the point $P$ ends up at $(-1,0)$. According to this new definition, $\cos 180^\circ$ becomes $-1$

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In a sense, there are two different notions of trigonometric functions — although they do agree with each other on their common domain, so to speak.

One concept is that of trigonometric functions of an acute angle in a right triangle. This definition ONLY makes sense for angles $0^{\circ}<\theta<90^{\circ}$, or $0<\theta<\frac{\pi}{2}$ in radians. There's no smallest or largest possible value of $\theta$ here (for example, $\theta$ can be an arbitrarily small positive number). But from this point of view, expressions like "$\cos(0^{\circ})$" or "$\cos(180^{\circ})$" certainly do NOT make any sense, because there are no such right triangles.

But then there's a much more general concept of trigonometric functions as functions defined for all real numbers. Geometrically, one possible way to introduce them is via the unit circle. With this definition, statements such as "$\cos(0^{\circ})=1$" or "$\cos(180^{\circ})=-1$" make perfect sense. And by the way, note that for angles lying within the first quadrant this definition coincides with the right triangle definition.

So the answer depends on the context. There are certainly no triangles with angles of $0^{\circ}$ or $180^{\circ}$. Whether that invalidates trig functions of such angles or not… see above.

zipirovich
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