How to Solve for r and θ in a Circle's Minor Arc and Sector Area Problem?

[ai93]

The length of the minor arc of a circle is 10cm, while the area of the sector AOB is 150cm2.

a) Form two equations involving r and θ, where θ is measured in radians.

b) Solve these equations simultaneously to find r and θ.

Help to solve? Cant understand the question very well.

I think the arc length formula was

length=$$\frac{n}{360}\cdot2\pi(r)$$

$$\therefore10=\frac{n}{360}\cdot2\pi(r)$$

The question states, Form two equations involving r and θ, where θ is measured in radians.

So would we have to arrange the formula to find the r and $$\theta$$

 

[MarkFL]

The formulas you want are:

Circular arc-length:

$$s=r\theta\tag{1}$$

Area of circular sector

$$A=\frac{1}{2}r^2\theta\tag{2}$$

Now, if we solve (1) for $\theta$, we find:

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

Now, substituting this into (2), we obtain:

$$A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{rs}{2}\implies r=\frac{2A}{s}\implies\theta=\frac{s^2}{2A}$$

Now, just plug in the given values for $A$ and $s$. :D

 

[ai93]

The formulas you want are:

Circular arc-length:

$$s=r\theta\tag{1}$$

Area of circular sector

$$A=\frac{1}{2}r^2\theta\tag{2}$$

Now, if we solve (1) for $\theta$, we find:

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

Now, substituting this into (2), we obtain:

$$A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{rs}{2}\implies r=\frac{2A}{s}\implies\theta=\frac{s^2}{2A}$$

Now, just plug in the given values for $A$ and $s$. :D

Im finding trouble understanding the question.
"Form two equations involving r and $$\theta$$" Basically means transpose the formula or make the equation equal to r and $$\theta$$?

Since the arc length = $$s=r\theta$$ (I thought arc length was length=$$\frac{n}{360}\cdot2\pi(r)$$?

r=$$s\theta$$?

For b) we just sub in the values? Where s is the length?

 

[MarkFL]

You are given values for the arc-length $s$ and the area $A$, and so you want to use the formulas relating $r$ and $\theta$ to these given values (which I gave as (1) and (2)) to be able to express both $r$ and $\theta$ as functions of $s$ and $A$ (which I did using some algebra), so that you can use these given values to determine $r$ and $\theta$.

Once you have $r$ and $\theta$ as functions of $s$ and $A$, it is simply a matter of using the given values to evaluate $r$ and $\theta$.

 

[SuperSonic4]

Im finding trouble understanding the question.
"Form two equations involving r and $$\theta$$" Basically means transpose the formula or make the equation equal to r and $$\theta$$?

It's asking to make two equations for the length of an arc and the area of a sector. As long as your equations have [math]r[/math] and [math]/theta[/math] in them then you're fine for this step.

MarkFL has given you the equations you need (his TeX is also better than mine! (Smile))$$\displaystyle s=r\theta\tag{1}$$

$$\displaystyle A=\frac{1}{2}r^2\theta\tag{2}$$

As you see both of these equations have [math]r[/math] and [math]\theta[/math] in them so Part A of the question has been solved already :)

Since the arc length = $$s=r\theta$$ (I thought arc length was length=$$\frac{n}{360}\cdot2\pi(r)$$?

r=$$s\theta$$?

$$\frac{n}{360}\cdot 2\pi r$$ is the equation for arc length where [math]n[/math] is in degrees. Since the question asks for radians you want $$s = r\theta$$. See the spoiler below for why they are equivalent

Spoiler

By definition a full circle has 360 degrees and $$2\pi$$ radians so you can say $$360^{\circ} = 2\pi$$ (we omit the ^c symbol for radians)

Your formula for degrees is $$s = \frac{\theta}{360} \cdot 2\pi r$$ but since we know that $$360^{\circ} = 2\pi$$ it can be subbed in

$$s = \frac{\theta}{2\pi} \cdot 2\pi r = r \theta$$ - the radian expression

For b) we just sub in the values? Where s is the length?

Sub in the values given (10cm and 150cm²) for s and A respectively so you have two equations and two unknowns ($$r$$ and $$\theta$$).

You can now solve simultaneously using either the substitution method MarkFL showed in post 2. He found $$\theta$$ in terms of [math]r[/math] which means he was able to substitute [math]\frac{s}{r}[/math] wherever [math]\theta[/math] appeared.

He did in terms of [math]s[/math] in post 2 but if you find it easier you may use [math]s=10[/math] for your example

 

[ai93]

It's asking to make two equations for the length of an arc and the area of a sector. As long as your equations have [math]r[/math] and [math]/theta[/math] in them then you're fine for this step.

MarkFL has given you the equations you need (his TeX is also better than mine! (Smile))$$\displaystyle s=r\theta\tag{1}$$

$$\displaystyle A=\frac{1}{2}r^2\theta\tag{2}$$

As you see both of these equations have [math]r[/math] and [math]\theta[/math] in them so Part A of the question has been solved already :)

$$\frac{n}{360}\cdot 2\pi r$$ is the equation for arc length where [math]n[/math] is in degrees. Since the question asks for radians you want $$s = r\theta$$. See the spoiler below for why they are equivalent

Spoiler

By definition a full circle has 360 degrees and $$2\pi$$ radians so you can say $$360^{\circ} = 2\pi$$ (we omit the ^c symbol for radians)

Your formula for degrees is $$s = \frac{\theta}{360} \cdot 2\pi r$$ but since we know that $$360^{\circ} = 2\pi$$ it can be subbed in

$$s = \frac{\theta}{2\pi} \cdot 2\pi r = r \theta$$ - the radian expression

Sub in the values given (10cm and 150cm²) for s and A respectively so you have two equations and two unknowns ($$r$$ and $$\theta$$).

You can now solve simultaneously using either the substitution method MarkFL showed in post 2. He found $$\theta$$ in terms of [math]r[/math] which means he was able to substitute [math]\frac{s}{r}[/math] wherever [math]\theta[/math] appeared.

He did in terms of [math]s[/math] in post 2 but if you find it easier you may use [math]s=10[/math] for your example

Thanks this really did help :D

 

[ai93]

The formulas you want are:

Circular arc-length:

$$s=r\theta\tag{1}$$

Area of circular sector

$$A=\frac{1}{2}r^2\theta\tag{2}$$

Now, if we solve (1) for $\theta$, we find:

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

Now, substituting this into (2), we obtain:

$$A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{rs}{2}\implies r=\frac{2A}{s}\implies\theta=\frac{s^2}{2A}$$

Now, just plug in the given values for $A$ and $s$. :D

Beginning to understand this a lot easier.
Just a question on the last part.
How did you rearrange to get $$\theta$$?

It went from, $$A=\frac{1}{2}r^2\left(\frac{s}{r}\right)$$ which I understand to $$=\frac{rs}{2}$$?

 

[MarkFL]

Beginning to understand this a lot easier.
Just a question on the last part.
How did you rearrange to get $$\theta$$?

$$s=r\theta$$

Divide through by $r$

$$\frac{s}{r}=\frac{\cancel{r}\theta}{\cancel{r}}$$

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

It went from, $$A=\frac{1}{2}r^2\left(\frac{s}{r}\right)$$ which I understand to $$=\frac{rs}{2}$$?

$$A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{\cancel{r}\cdot r\cdot s}{2\cancel{r}}=\frac{rs}{2}$$