# Thread: Arc Length and Chord Length

1. ## 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

2. 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.

3. 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 $\displaystyle \displaystyle \theta$.

4. 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?

5. 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?

6. 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...

7. 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.

8. ok, I understand now

Thank you

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