# Central angle of a circular sector from area and arc length

I've been doing a task which says the following:

Area of a circular sector is $3.2\pi cm^2$, arc length is $0.8\pi cm$. What is the central angle?

I've been attacking this from several angles (boo), and finally settled on approaching this from area aspect of it. This is what I did:

$A = \frac{rl}{2}$ -> $r=\frac{2A}{l}$, where r is radius and l arc length.

So then, I did:

$r=\frac{2*3.2\pi}{0.8\pi}$->$\frac{6.4\pi}{0.8\pi}$->$\frac{\frac{32\pi}{5}}{\frac{4\pi}{5}}$->$\frac{32\pi}{5}*\frac{5}{4\pi}=8$

From there, I thought arc length formula was appropriate to rearrange into theta.

$s=r\theta$->$\theta=\frac{s}{r}$

and then I did it with numbers:

$\theta=\frac{0.8\pi}{8}=\frac{8\pi}{10}*\frac{10}{80}=\frac{\pi}{10}$

$\frac{\pi}{10}$ radians, hopefully. Which should be $\frac{180}{\pi}*\frac{\pi}{10}=18°$

Problem now is that since I have no solutions provided in this textbook, I can't check the answer by looking at one, so I opted to check it back by plugging in newly found radius and arc length to find if the area matches with given one.

$A=\frac{\frac{8}{1}*\frac{\pi}{10}}{2}=\frac{4\pi}{5}cm^2$ which isn't right.

I am sure that either I'm making a stupid mistake somewhere with units or my approach to this is completely wrong, but I am stuck.

You got the right value for ${\theta}$, but your radius should be squared in the formula for area of sector.
The area of a circle is ${\pi}r^2$.
If ${\theta}$ is in radians, then the area of a sector is ${\pi}r^2$*($\frac{\theta}{2{\pi}}$)
So we have Area of sector = $\frac{r^2\theta}{2}$
So, plugging in r=8 and ${\theta}={\frac{\pi}{10}}$ gets you the required value of 3.2${\pi}cm^2$
• I see. So, I've made a same mistake in the beginning as well where $A = \frac{rl}{2}$ -> $r=\frac{2A}{l}$, where r is radius and l arc length should have actually been $A = \frac{r^2l}{2}$ -> $r^2=\frac{2A}{l}$. – MathAgain Mar 30 '15 at 22:21
• Well no- I think that was correct in your beginning statement. ${l}$ = r${\theta}$, so $\theta = \frac{l}{r}$ Substituting into A=$\frac{r^2{\theta}}{2}$ yields your original formula: A = $\frac{r{l}}{2}$ – Mike Andrejkovics Mar 30 '15 at 22:59