have trouble with this limit question 
a) By considering the areas of the triangle OAD, the sector OAC and the triangle OBC,
show that
$(\cos \theta)(\sin \theta) < \theta < \frac{\sin\theta}{\cos\theta}$
I find out:
Area of OAD=$\frac{1}{2}OD\cdot AD \cdot \sin \theta$
Area of OAC=$\frac{1}{2}OC^2 \theta$
Area of OBC=$\frac{1}{2}OC\cdot BC \cdot\sin\theta$
Now I'm stuck at how to apply this to prove
How to prove?
(b) Use (a) and the Squeeze Theorem to show that
$\displaystyle\lim_{\theta\to 0^+}\frac{\sin\theta}{\theta}= 1$
 A: Hint: WORK IN RADIANS!
$a)$ $$\text{Area of }\Delta OAD=\dfrac{1}{2}\cdot OD\cdot AD\\
\text{Area of sector }OAC=\dfrac{\theta}{360}\pi (OA)^2\\
\text{Area of }\Delta OBC=\dfrac{1}{2}\cdot OC\cdot BC$$
See that
$$\text{Area of }\Delta OAD<\text{Area of sector }OAC<\text{Area of }\Delta OBC\\
\implies \dfrac{1}{2}\cdot \cos\theta\cdot \sin\theta<\dfrac{\theta}{360}\pi (OA)^2<\dfrac{1}{2}\cdot 1\cdot BC$$
Now,
$$DC=1-\cos\theta\\
BC=\tan\theta$$
$b)$ Then, after doing $a)$, use the fact that
$$\dfrac{1}{2}\sin\theta\cos\theta<\theta/2\\
\theta/2<\dfrac{1}{2}\tan\theta$$
Then use the squeeze theorem. The limit follows.
A: The geometry has been discussed in comments. So we have 
$$\frac{1}{2}\cos\theta\lt \frac{1}{2}\theta.$$
Multiply through by $2$. We get $\cos\theta\sin\theta\lt \theta$.
Divide both sides by $\theta\cos\theta$. We get
$$\frac{\sin\theta}{\theta}\lt \frac{1}{\cos\theta}.\tag{1}$$
We also got from the geometry that
$$\frac{1}{2}\theta\lt \frac{1}{2}\frac{\sin \theta}{\cos\theta}.$$ Multiplying by $2$, and rearranging, we get $\theta\cos\theta\lt \sin\theta$. Divide through by $\theta$. We get
$$\cos\theta\lt\frac{\sin\theta}{\theta}.\tag{2}$$
Now putting (1) and (2) together, we get
$$\cos\theta\lt \frac{\sin\theta}{\theta}\lt \frac{1}{\cos\theta}.\tag{3}$$
Finally, let $\theta\to 0$. The left wall and the right wall in (3) both approach $1$, and poor $\frac{\sin\theta}{\theta}$ is therefore forced to approach $1$. 
A: by the scheme we have that the area of triangle $OAD$ is $$A_1=\frac{\cos \theta\cdot \sin \theta}{2},$$
The area os sector $OAC$ is 
$$A_2=\frac{1}{2}R^2\theta=\frac{\theta}{2}, $$
because $R=1$.
The area of triangle $OCB$ is
$$A_3=\frac{1\cdot \tan \theta}{2}=\frac{\tan \theta}{2}. $$
Now, note that $A_1\leq A_2\leq A_2$, i.e.,
$$\frac{\sin \theta\cdot \cos \theta}{2}\leq \frac{\theta}{2}\leq \frac{\tan \theta}{2}, $$
i.e.,
$$\sin \theta\cdot \cos \theta \leq \theta \leq \frac{\sin \theta}{\cos \theta}.  $$
This answers the item (a).
For (b): dividing the inequality in (a) by $\sin \theta$ (which is positive because $0<\theta<\frac{\pi}{2}$,c.f. picture) we obtain
$$\cos \theta\leq \frac{\theta}{\sin \theta} \leq \frac{1}{\cos \theta}$$
and "inverting", we obtain
$$\cos \theta\leq \frac{\sin\theta}{\theta}\leq \frac{1}{\cos \theta} $$
By the theorem of "sandwich", passing the limit when $\theta\rightarrow 0^+$ we obtain that
$$1\leq\lim_{\theta\rightarrow 0^+}\frac{\sin \theta}{\theta}\leq 1, $$
i.e., 
$$\lim_{\theta\rightarrow 0^+}\frac{\sin \theta}{\theta}= 1. $$
