• 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 = $\theta$ radians. The length of the arc PQ is 6 cm and the area of the sector PQ is $22.5cm^2$.
Find the values of r and $\theta$. [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 = $\theta$ radians. The length of the arc PQ is 6 cm and the area of the sector PQ is $22.5cm^2$.
Find the values of r and $\theta$. [5]

Any help towards answering this is greatly appreciated.

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

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

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

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

evaluate $r$, then go back to either of the original equations and solve for $\theta$.
• 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 = $\theta$ radians. The length of the arc PQ is 6 cm and the area of the sector PQ is $22.5cm^2$.
Find the values of r and $\theta$. [5]

Any help towards answering this is greatly appreciated.

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

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

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

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

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

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

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

$3r = 22.5$

$r = 67.5\textrm{cm}$.

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

Originally Posted by Prove It

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

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

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

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

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

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

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

$3r = 22.5$

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

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

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

Originally Posted by skeeter
.

Oops... Duh...

I was just testing you :P