1. ## [SOLVED] Trignometry and Radians Question

Question:

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

(i) Show that the area, $A\ cm^2$, of the shaded region $AXB$ is given by
$A = \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)$
(ii) In the case where $r = 12$ and $\theta = \frac{1}{6}\pi$, find the perimeter of the shaded region $AXB$, leaving your answer in terms of $\sqrt{3}$ and $\pi$.

Attempt:

(i) Area of shaded region = Area of sector of circle - Area of triangle
Area of sector of circle $= \frac{1}{2}r^2\theta$

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

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

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

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

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

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

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

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

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

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

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

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

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 $= r\theta$

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

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

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

2. Originally Posted by looi76
Question:

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

(i) Show that the area, $A\ cm^2$, of the shaded region $AXB$ is given by
$A = \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)$
(ii) In the case where $r = 12$ and $\theta = \frac{1}{6}\pi$, find the perimeter of the shaded region $AXB$, leaving your answer in terms of $\sqrt{3}$ and $\pi$.

Attempt:

(i) Area of shaded region = Area of sector of circle - Area of triangle
Area of sector of circle $= \frac{1}{2}r^2\theta$

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

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

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

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

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

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

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

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

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

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

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

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

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 $= r\theta$

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

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

How can I get the length of $AX$ in terms of $\pi$? I know that $\sin{30^o} = \frac{1}{2}$ but if I used this would the $\pi$ vanish?
For part 1: Aside from calling your triangle a circle a few times, the work for the formula is correct.

For part 2: The length of AX is not going to be in terms of pi. It is simply $r \sin{(\theta)}$, which, if $\theta = \frac{\pi}{6}$, is equal to 6.

3. Hi again !

Originally Posted by looi76
are my steps for showing the area of the shaded region correct?
Well, I find it correct and well-explained

Originally Posted by looi76
(ii) Perimeter of Shaded Region = Arc lenght of circle AB + AX + XB

Arc length of circle $= r\theta$

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

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

How can I get the length of $AX$ in terms of $\pi$? I know that $\sin{30^o} = \frac{1}{2}$ but if I used this would the $\pi$ vanish?[/SIZE][/FONT]
There is still $\pi$ due to the arc length of circle, isn't it ? You don't necessarily have $\pi$ in AX.

And what about XB ? ( $\sqrt{3}$ will be used here)

4. Do you know the formula $\frac{1}{2}ab\sin C$ for the area of a triangle.

It makes this question much simpler.
One side of the triangle is r, another is $r\cos \theta$...

5. Originally Posted by a tutor
Do you know the formula $\frac{1}{2}ab\sin C$ for the area of a triangle.

It makes this question much simpler.
One side of the triangle is r, another is $r\cos \theta$...
Thanks for reminding me. HOW DID I FORGET THAT?