# Arc Length and Chord Length

• Dec 20th 2010, 01:19 AM
jgv115
Arc Length and Chord Length
Find the magnitude, in degrees and minutes, of the angle subtended at the centre of a circle of radius length 30 cm, by a chord of length 50 cm.

I know that the formula to work out the chord length is $\displaystyle 2r\sin\frac{\theta}{2}$

$\displaystyle \theta$ is radians

So I get

$\displaystyle \frac{5}{6} =\sin\frac{\theta}{2}$

What do I do now?

I'm new to all this radian stuff (I'm completely clueless). I only know how to convert degrees to radians and vice versa
• Dec 20th 2010, 04:11 AM
Prove It
If you draw this, you get an isosceles triangle, with two sides = 30cm long and the third side = 50cm long.

Since you know three sides and are trying to find an angle, you can use the cosine rule.

$\displaystyle \displaystyle \cos{C} = \frac{a^2 + b^2 - c^2}{2ab}$

$\displaystyle \displaystyle \cos{C} = \frac{30^2 + 30^2 - 50^2}{2(30)(30)}$

$\displaystyle \displaystyle \cos{C} = \frac{900 + 900 - 2500}{1800}$

$\displaystyle \displaystyle \cos{C} = -\frac{700}{1800}$

$\displaystyle \displaystyle \cos{C} = -\frac{7}{18}$

$\displaystyle \displaystyle C = \cos^{-1}{\left(-\frac{7}{18}\right)}$.

Put this into your calculator, making sure it is in degree mode, and convert to DMS.
• Dec 20th 2010, 12:05 PM
SammyS
Quote:

Originally Posted by jgv115
Find the magnitude, in degrees and minutes, of the angle subtended at the centre of a circle of radius length 30 cm, by a chord of length 50 cm.

I know that the formula to work out the chord length is $\displaystyle 2r\sin\frac{\theta}{2}$

$\displaystyle \theta$ is radians

So I get

$\displaystyle \frac{5}{6} =\sin\frac{\theta}{2}$

What do I do now?

I'm new to all this radian stuff (I'm completely clueless). I only know how to convert degrees to radians and vice versa

$\displaystyle \displaystyle \text{chord length }=2r\sin\frac{\theta}{2}$ no matter what units you use for $\displaystyle \displaystyle \theta$.

Solve: $\displaystyle \displaystyle \frac{5}{6} =\sin\frac{\theta}{2}$ by taking $\displaystyle \displaystyle \sin^{-1}$ of both sides, giving you:

$\displaystyle \displaystyle \sin^{-1}\left( \frac{5}{6} \right) = {{\theta}\over{2}}$.

Multiply both sides by 2 to find $\displaystyle \displaystyle \theta$.

• Dec 20th 2010, 03:47 PM
jgv115
Alright, thank you.

A chord of length 6 cm is drawn in a circle of radius 7 cm. Find the area of the smaller region inside the circle cut off by the chord.

The book says the formula is $\displaystyle \frac {\pi r^2\theta}{360} - \frac{1}{2} r^2 \sin \theta$

But then they go on to use radians:

They do

$\displaystyle \phi = \frac {\pi\theta}{180}$

So the formula becomes $\displaystyle \frac{1}{2}r^2(\phi-\sin\phi})$ After factorising

If I use the first formula:

$\displaystyle \frac {\pi r^2\theta}{360} - \frac{1}{2} 7^2 \sin 50.7359$ (I already worked out the angle)

$\displaystyle \displaytype \frac {\pi*7^2*50.7539}{360}-\frac{1}{2}r^2\sin\theta$

$\displaystyle = 2.73$

Which is correct

$\displaystyle \frac{1}{2}r^2(\phi-\sin\phi})$

$\displaystyle \displaytype \frac{1}{2}*49(\frac{50.7539*\pi}{180}-\sin\frac{50.7539*\pi}{180})$

I get like 21.3239 for some reason.

Am I doing anything wrong?
• Dec 20th 2010, 04:37 PM
Quote:

Originally Posted by jgv115
Alright, thank you.

A chord of length 6 cm is drawn in a circle of radius 7 cm. Find the area of the smaller region inside the circle cut off by the chord.

The book says the formula is $\displaystyle \frac {\pi r^2\theta}{360} - \frac{1}{2} r^2 \sin \theta$

But then they go on to use radians:

They do

$\displaystyle \phi = \frac {\pi\theta}{180}$

So the formula becomes $\displaystyle \frac{1}{2}r^2(\phi-\sin\phi})$ After factorising

If I use the first formula:

$\displaystyle \frac {\pi r^2\theta}{360} - \frac{1}{2} 7^2 \sin 50.7359$ (I already worked out the angle)

$\displaystyle \displaytype \frac {\pi*7^2*50.7539}{360}-\frac{1}{2}r^2\sin\theta$

$\displaystyle = 2.73$

Which is correct

$\displaystyle \frac{1}{2}r^2(\phi-\sin\phi})$

$\displaystyle \displaytype \frac{1}{2}*49(\frac{50.7539*\pi}{180}-\sin\frac{50.7539*\pi}{180})$

I get like 21.3239 for some reason.

Am I doing anything wrong?

• Dec 20th 2010, 05:55 PM
jgv115
Ok I got it! I worked out I had to type the degrees sign and the calculator will automatically convert it to radians.

Haha! I'm getting there...
• Dec 20th 2010, 06:59 PM
Prove It
Quote:

Originally Posted by jgv115
Alright, thank you.

A chord of length 6 cm is drawn in a circle of radius 7 cm. Find the area of the smaller region inside the circle cut off by the chord.

The book says the formula is $\displaystyle \frac {\pi r^2\theta}{360} - \frac{1}{2} r^2 \sin \theta$

But then they go on to use radians:

They do

$\displaystyle \phi = \frac {\pi\theta}{180}$

So the formula becomes $\displaystyle \frac{1}{2}r^2(\phi-\sin\phi})$ After factorising

If I use the first formula:

$\displaystyle \frac {\pi r^2\theta}{360} - \frac{1}{2} 7^2 \sin 50.7359$ (I already worked out the angle)

$\displaystyle \displaytype \frac {\pi*7^2*50.7539}{360}-\frac{1}{2}r^2\sin\theta$

$\displaystyle = 2.73$

Which is correct

$\displaystyle \frac{1}{2}r^2(\phi-\sin\phi})$

$\displaystyle \displaytype \frac{1}{2}*49(\frac{50.7539*\pi}{180}-\sin\frac{50.7539*\pi}{180})$

I get like 21.3239 for some reason.

Am I doing anything wrong?

Again, draw the circle with the chord. Connect each end of the chord to the centre. You should find you have a triangle of lengths 7cm, 7cm, 6cm.

Use Heron's Formula to find the area of this triangle.

Use the Cosine Rule to find the angle between the two radii.

Use $\displaystyle \displaystyle \frac{r^2 \theta}{2}$ to find the area of the sector.

Subtract the area of the triangle from the area of the sector.
• Dec 22nd 2010, 01:24 AM
jgv115
ok, I understand now

Thank you