[SOLVED] Trignometry and Radians Question

**Question:**

In the diagram, $\displaystyle AB$ is an arc of a circle, centre $\displaystyle O$ and radius $\displaystyle r \ cm$, and angle $\displaystyle AOB = \theta \ radians$. The point X lies on OB and AX is perpendicular OB.

(i) Show that the area, $\displaystyle A\ cm^2$, of the shaded region $\displaystyle AXB$ is given by

$\displaystyle A = \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)$

(ii) In the case where $\displaystyle r = 12 $ and $\displaystyle \theta = \frac{1}{6}\pi$, find the perimeter of the shaded region $\displaystyle AXB$, leaving your answer in terms of $\displaystyle \sqrt{3}$ and $\displaystyle \pi$.

**Attempt:**

(i) Area of shaded region = Area of sector of circle - Area of triangle

Area of sector of circle $\displaystyle = \frac{1}{2}r^2\theta$

Area of Triangle$\displaystyle = \frac{1}{2} \times Lenght \times Breath$

Length $\displaystyle \rightarrow \sin\theta = \frac{AX}{r}$

Length $\displaystyle \rightarrow AX = \sin\theta \times r$

Length $\displaystyle \rightarrow AX = \sin\theta{r}$

Breath $\displaystyle \rightarrow \cos\theta = \frac{OX}{r}$

Breath $\displaystyle \rightarrow OX = \cos\theta \times r$

Breath $\displaystyle \rightarrow OX = \cos\theta{r}$

Area of Triangle$\displaystyle = \frac{1}{2} \times \sin\theta{r} \times \cos\theta{r}$

Area of Triangle$\displaystyle = \frac{1}{2}r^2\sin\theta\cos\theta$

Area of shaded region = Area of sector of circle - Area of triangle

$\displaystyle A \ = \ \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\cos\theta$

$\displaystyle A \ = \ \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)$

are my steps for showing the area of the shaded region correct?

(ii) Perimeter of Shaded Region = Arc lenght of circle AB + AX + XB

Arc length of circle $\displaystyle = r\theta$

Arc length of circle $\displaystyle = 12 \times \frac{1}{6}\pi$

Length of AX $\displaystyle = \sin{\frac{1}{6}\pi} \times 12$

How can I get the length of $\displaystyle AX$ in terms of $\displaystyle \pi$? I know that $\displaystyle \sin{30^o} = \frac{1}{2}$ but if I used this would the $\displaystyle \pi$ vanish?(Speechless)