How does the way we define cos or tan have anything to do with degrees of the angle?

Question

So sine of angle $A$ is just a ratio. It is the ratio of the length of the opposite or perpendicular of angle $A$ and the hypotenuse.
Cosine of angle $A$ is also just a ratio. It is the ratio of the length of base of $A$ and the hypotenuse.
Tangent of angle $A$ is the ratio of the length of the perpendicular to $A$ and length of the base of $A$.
So far so good. These are just ratios.
Sine of $A$ makes sense since it determines how big angle $A$ is.
But the cosine of angle $A$ or tangent of $A$ is not intuitive to me.
For example in what way the length of base of $A$ and the hypotenuse's length affect the angle of $A$? Their length doesn't seem relevant to how many degrees it has. Same for tangent.
Can someone please help me on this?

Answer 1

Here’s a somewhat different way to define these functions that might make the relationships to sides of triangles clearer.

Consider the unit circle with center $O$ and two rays from $O$ that intersect the circle at points $A$ and $B$, with an acute angle $\theta=\angle AOB$ between them, as shown below.

Instead of defining the trigonometric functions as ratios of sides of triangles, define them as the lengths of certain line segments as follows:

• $\sin\theta = \overline{BC}$, the altitude to $\overrightarrow{OA}$ from $B$.
• $\tan\theta=\overline{AD}$, the length of the segment tangent to the circle at $A$ that intersects $\overrightarrow{OB}$ at $D$.
• $\sec\theta=\overline{OD}$, the distance to the intersection of $\overrightarrow{OB}$ with the tangent to the circle at $A$.

The “co” functions are the lengths of the corresponding line segments drawn on the other side of $\overrightarrow{OB}$, using the ray $\overrightarrow{OE}$, which is perpendicular to $\overrightarrow{OA}$.

$\triangle{OCB}$ is congruent to $\triangle{BFO}$, so $\overline{BF}=\overline{OC}$. Moreover, $\overline{OB}=1$, so we have the familiar ratios of sides of a right triangle for the sine and cosine.

Now, $\triangle{OCB}$ and $\triangle{OAD}$ are similar, so $\overline{AD}:\overline{BC}::\overline{OA}:\overline{OC}$ and $\overline{OA}=1$, and we have $$\tan\theta=\overline{AD}={\overline{BC}\over\overline{OC}}={\sin\theta\over\cos\theta}.$$ From these same triangles and the previous equation we find that $$\sec\theta = {\tan\theta\over\sin\theta} = \frac1{\cos\theta}.$$

The remaining two functions can be related to $\triangle{OCB}$ and the other functions via similar considerations.

If we change the radius of the circle, it’s clear that the lengths of all of these line segments change proportionally, i.e., the ratios of their lengths to the radius are constant across similar triangles. Since we’re taking a radius as the hypotenuse of the triangle, this also means that their ratios to the hypotenuse are constant for a given angle, which leads to the more familiar definitions of these functions as ratios of sides of a right triangle.

Answer 2

Every right angle with have distinct sides a, b, and hypotenuse h, always with the condition $a^2 + b^2 = h^2$. Each triangle with an (a,b,h) sides will have a distinct angle with with a one-to-one coorespondence between angle $\theta$ and the set of $(a/h, b/h, 1)$ where $(a/h, b/h, 1)$ represents a class of similar right triangles-- similar up to a scaling factor.

$a/h$ is unique to the angle. We can that $sin$. $b/h$ is unique to the angle we call that $cos$. $a/b = (a/h)/(b/h)$ is also unique. That is $tan$.

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You say you understand that $A/H$ determines how big the angle is. But $B/H = \sqrt{1 - A/H}$ by the pythagorean theorem. If one determines how big the angle is, the other has to also determine how big it is.

Also if you just flip the triangle sideways, the height becomes the base, and the base becomes the height. So whatever was true about the height of one must be true about the base of the other.

Now if $A/H$ determines how big the angle is, so must $\frac{A/H}{\sqrt{1 - A/H}} = \frac{A/H}{B/H} = A/B = \tan$.