Question about the proof of a limit Here is the picture in question:

In the proof of $$\lim_{x\to 0}\frac {\sin x}{x} = 1$$ involving the unit circle, how did they get the height of the largest right triangle to be $\tan(x) $?  Shouldn't it be equivalent to $\sin(x) $?  I feel as if the answer to this question is really obvious, but I've tried solving it by substitution of opposite, adjacent, etc. and it just doesn't work out.
Thanks!
 A: If the radius of the circle is 1, then the triangle involving $\tan(x)$ is one whose base is 1 and whose height is some number, say $h$, and whose angle is given by $x$.
In such a case, the "adjacent" side has length 1, and the "opposite" side has length $h$. Thus
$$
\tan(x) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{1} = h
$$
so the unknown height is given by $\tan(x)$ as claimed.
A: Hint: if you look at the two similars triangles $\hat{ACH}$ and $\hat{ADB}$, you have $$\frac{CH}{AH}=\frac{DB}{AB}$$
$$\frac{\sin(x)}{\cos(x)}=\tan(x)=DB$$
denoting $H$ the projection of $C$ on $AB$.
A: As the angle goes up to a right angle, notice that the thing in the picture labeled $\sin x$ goes up only as far as $1$, whereas the thing labeled $\tan x$ goes up to $\infty$.  That should tell you that it certainly cannot be $\sin x$.
It's the tangent because $\tan$ is opposite over adjacent, and the length of the adjacent side is $1$.
A: Call the height $H$. Then $\tan x = {\rm opp}/{\rm adj} = H/1 = H$.
