• Mar 12th 2009, 04:11 PM
db5vry
This question is something to do with radian measure. I've never studied this before (I understand a few of the formulas generally used with no idea how they are applied) and want to solve this question:

Two points P and Q lie on a circle with centre O. The radius of the circle is r cm and angle POQ = $\displaystyle \theta$ radians. The length of the arc PQ is 6 cm and the area of the sector PQ is $\displaystyle 22.5cm^2$.
Find the values of r and $\displaystyle \theta$. [5]

Any help towards answering this is greatly appreciated.
• Mar 12th 2009, 04:29 PM
skeeter
Quote:

Originally Posted by db5vry
This question is something to do with radian measure. I've never studied this before (I understand a few of the formulas generally used with no idea how they are applied) and want to solve this question:

Two points P and Q lie on a circle with centre O. The radius of the circle is r cm and angle POQ = $\displaystyle \theta$ radians. The length of the arc PQ is 6 cm and the area of the sector PQ is $\displaystyle 22.5cm^2$.
Find the values of r and $\displaystyle \theta$. [5]

Any help towards answering this is greatly appreciated.

arc length, $\displaystyle s = r\theta$ , $\displaystyle \theta$ is in radians

sector area, $\displaystyle A = \frac{r^2 \theta}{2}$ , $\displaystyle \theta$ in radians

$\displaystyle A = \frac{r^2 \theta}{2} = \frac{r \cdot r\theta}{2} = \frac{r \cdot s}{2}$

$\displaystyle r = \frac{2A}{s}$

evaluate $\displaystyle r$, then go back to either of the original equations and solve for $\displaystyle \theta$.
• Mar 12th 2009, 05:09 PM
Prove It
Quote:

Originally Posted by db5vry
This question is something to do with radian measure. I've never studied this before (I understand a few of the formulas generally used with no idea how they are applied) and want to solve this question:

Two points P and Q lie on a circle with centre O. The radius of the circle is r cm and angle POQ = $\displaystyle \theta$ radians. The length of the arc PQ is 6 cm and the area of the sector PQ is $\displaystyle 22.5cm^2$.
Find the values of r and $\displaystyle \theta$. [5]

Any help towards answering this is greatly appreciated.

You're told the radius is $\displaystyle r$ and that the length of the arc $\displaystyle PQ$ is $\displaystyle 6\textrm{bf}$.

Therefore the angle $\displaystyle \theta$, in radians, is $\displaystyle \frac{6}{r}$.

The area of a sector, if the angle is given in radians, is

$\displaystyle \frac{\theta}{2\pi}\pi r^2 = \frac{\theta}{2}r^2$.

We know that $\displaystyle \theta = \frac{6}{r}$ and that the area is $\displaystyle 22.5\textrm{cm}^2$.

So $\displaystyle \frac{\frac{6}{r}}{2}r^2 = 22.5$

$\displaystyle \frac{3}{r}r^2 = 22.5$

$\displaystyle 3r = 22.5$

$\displaystyle r = 67.5\textrm{cm}$.

From this, we can see $\displaystyle \theta = \frac{6}{67.5}^C$.
• Mar 12th 2009, 05:13 PM
skeeter
Quote:

Originally Posted by Prove It

You're told the radius is $\displaystyle r$ and that the length of the arc $\displaystyle PQ$ is $\displaystyle 6\textrm{bf}$.

Therefore the angle $\displaystyle \theta$, in radians, is $\displaystyle \frac{6}{r}$.

The area of a sector, if the angle is given in radians, is

$\displaystyle \frac{\theta}{2\pi}\pi r^2 = \frac{\theta}{2}r^2$.

We know that $\displaystyle \theta = \frac{6}{r}$ and that the area is $\displaystyle 22.5\textrm{cm}^2$.

So $\displaystyle \frac{\frac{6}{r}}{2}r^2 = 22.5$

$\displaystyle \frac{3}{r}r^2 = 22.5$

$\displaystyle 3r = 22.5$

$\displaystyle r = 67.5\textrm{cm}$. correction, r = 7.5 cm

From this, we can see $\displaystyle \theta = \frac{6}{67.5}^C$.

.
• Mar 14th 2009, 06:49 PM
Prove It
Quote:

Originally Posted by skeeter
.

Oops... Duh...

I was just testing you :P