Solve the problem that involves a sector of a circle

[chwala]

Homework Statement

See attached

Relevant Equations

Area of circle

Wawawawawawawawa part (a) despite being easy had me running to and fro for some time! Anyway;

Here we have;
$\dfrac{1}{2}r^2θ=\sin θ+\dfrac{3}{2}\sin θ$
$2θ=2.5\sin θ$

the required result follows...

I would appreciate an alternative approach to this...

For part (b), we have
$θ_2=1.25\sin θ_1$
$θ_2=1.25\sin 0.5=0.59928$

$θ_3=1.25\sin θ_2$
$θ_3=1.25\sin (0.59928)=0.70506$

$θ_4=1.25\sin θ_3$
$θ_4=1.25\sin (0.70506)=0.810$

...

 

[Delta2]

for a)
I can't be sure what exactly you had in mind to write but what you actually wrote isn't 100% correct.

The term on the LHS is correct (the $\frac{1}{2}r^2\theta$) but the two terms on the RHS ($\sin\theta+\frac{3}{2}\sin\theta$ are missing something. Probably a typo ok since it gets simplified in the next step.

It would help to mention the so called in between steps which you probably consider to be self implied so you omit them.
For example you use the formula $\frac{1}{2}(OB)(OM)\sin\hat O$ for the area of the triangle OMB, and to find the area of the MBA you use the given area ratio.

 

[chwala]

for a)
I can't be sure what exactly you had in mind to write but what you actually wrote isn't 100% correct.

The term on the LHS is correct (the $\frac{1}{2}r^2\theta$) but the two terms on the RHS ($\sin\theta+\frac{3}{2}\sin\theta$ are missing something. Probably a typo ok since it gets simplified in the next step.

It would help to mention the so called in between steps which you probably consider to be self implied so you omit them.
For example you use the formula $\frac{1}{2}(OB)(OM)\sin\hat O$ for the area of the triangle OMB, and to find the area of the MBA you use the given area ratio.

Arrrgh just a minute I check...

 

[chwala]

Homework Statement:: See attached
Relevant Equations:: Area of circle

View attachment 302862Wawawawawawawawa part (a) despite being easy had me running to and fro for some time! Anyway;

Here we have;
$\dfrac{1}{2}r^2θ=\sin θ+\dfrac{3}{2}\sin θ$
$2θ=2.5\sin θ$

the required result follows...

I would appreciate an alternative approach to this...

For part (b), we have
$θ_2=1.25\sin θ_1$
$θ_2=1.25\sin 0.5=0.59928$

$θ_3=1.25\sin θ_2$
$θ_3=1.25\sin (0.59928)=0.70506$

$θ_4=1.25\sin θ_3$
$θ_4=1.25\sin (0.70506)=0.810$

...

$\dfrac{1}{2}r^2θ=\dfrac{1}{2}×1×2× \sinθ +\dfrac{3}{2}\sin θ$
$\dfrac{1}{2}2^2θ=\sin θ+\dfrac{3}{2}\sin θ$
$2θ=2.5\sin θ$

Area of $MAB$=$\dfrac{3}{2}$ × Area of $OMB$.

This should be clear now. Cheers!

 

[Delta2]

Ehm it is not given anywhere that the radius of the circle is 2 or 4 units. and on the LHS you use r=4 while on the RHS seems you use r=2.

 

[chwala]

Ehm it is not given anywhere that the radius of the circle is 2 or 4 units. and on the LHS you use r=4 while on the RHS seems you use r=2.

Amended...just check again. We are told $M$ is midpoint...i therefore used the fact $1:1$ for $OM:MA$ and for subsequent working to solution.

 

[Delta2]

Amended...just check again.

Yes ok now it makes more sense but it isn't given anywhere in the statement of the problem that r=2. The full correct equation would be $$\frac{1}{2}r^2\theta=(1+\frac{3}{2})\frac{1}{2}r\frac{r}{2}\sin\theta$$ and then $\frac{1}{2}r^2$ gets simplified and we are left with $$\theta=2.5\frac{1}{2}\sin\theta$$

 

[chwala]

Yes ok now it makes more sense but it isn't given anywhere in the statement of the problem that r=2. The full correct equation would be $$\frac{1}{2}r^2\theta=(1+\frac{3}{2})\frac{1}{2}r\frac{r}{2}\sin\theta$$ and then $\frac{1}{2}r^2$ gets simplified and we are left with $$\theta=2.5\frac{1}{2}\sin\theta$$

My working is $100\%$ correct! we can use $1:1$ to come up with the linear scale values for the radius.

Where;

$OM$+ $MA$=Radius.

even if we were to use $50:50$, we would end up with;

$1250θ=625\sin θ+ 937.5\sin θ$

$θ=1.25\sin θ$

In essence, any ratio of form $x:x$ with radius =$2x$ will work.

 

[Delta2]

What you saying is correct but I don't understand why you have to replace $r$ by some specific value, let it be 50 or 100...

 

[chwala]

What you saying is correct but I don't understand why you have to replace $r$ by some specific value, let it be 50 or 100...

I agree with you...I was just trying to state that it's possible to come up with values for the radius and work to the solution...you questioned that...of course your approach in post $7$ is straightforward. Cheers mate.

 

[Delta2]

Yes well ok you can use any value for r BECAUSE the way algebra works, r gets simplified from the equation. But you got to prove that , that r gets simplified.

 

[Mark44]

Yes well ok you can use any value for r BECAUSE the way algebra works, r gets simplified from the equation. But you got to prove that , that r gets simplified.

I agree. Although it appears that the radius is 4 because of the tick marks, there is nothing in the drawing to show the scale. It's better to use R for the circle's radius than to assign a value arbitrarily.

Area(sector OAB) = $\frac \theta 2 R^2$.
From the given information we can infer that Area(ΔOMB) = 2/5 * Area (sector OAB) = $\frac 2 5 \frac \theta 2 R^2 = \frac \theta 5 R^2$.
On the other hand, using the formula for the area of a triangle, Area(ΔOMB) = $\frac 1 2 \frac R 2 R \sin(\theta) = \frac {R^2} 4 \sin(\theta)$.

Setting the last two areas equal, we get $\frac \theta 5 R^2 = \frac {R^2} 4 \sin(\theta) \Rightarrow \theta = \frac 5 4 \theta$, which was to be shown.

 

[chwala]

I agree. Although it appears that the radius is 4 because of the tick marks, there is nothing in the drawing to show the scale. It's better to use R for the circle's radius than to assign a value arbitrarily.

Area(sector OAB) = $\frac \theta 2 R^2$.
From the given information we can infer that Area(ΔOMB) = 2/5 * Area (sector OAB) = $\frac 2 5 \frac \theta 2 R^2 = \frac \theta 5 R^2$.
On the other hand, using the formula for the area of a triangle, Area(ΔOMB) = $\frac 1 2 \frac R 2 R \sin(\theta) = \frac {R^2} 4 \sin(\theta)$.

Setting the last two areas equal, we get $\frac \theta 5 R^2 = \frac {R^2} 4 \sin(\theta) \Rightarrow \theta = \frac 5 4 \theta$, which was to be shown.

Looks good...in this approach, $AMB$ was kept in the dark!

 

[chwala]

Yes well ok you can use any value for r BECAUSE the way algebra works, r gets simplified from the equation. But you got to prove that , that r gets simplified.

Just looking at this again, true there really was no need to assign values for $r$, my approach without $r$ values should realize the required solution.

 

[chwala]

$\dfrac{1}{2}r^2θ=\dfrac{1}{2}×1×2× \sinθ +\dfrac{3}{2}\sin θ$
$\dfrac{1}{2}2^2θ=\sin θ+\dfrac{3}{2}\sin θ$
$2θ=2.5\sin θ$

Area of $MAB$=$\dfrac{3}{2}$ × Area of $OMB$.

This should be clear now. Cheers!

$\dfrac{1}{2}r^2θ=\dfrac{1}{2}×\dfrac{1}{2}×r^2 \sinθ +\dfrac{3}{2} \left[\dfrac{1}{2}×\dfrac{1}{2}×r^2 \sinθ\right]$
$4θ =2\sinθ+3\sinθ$
$4θ=5\sinθ$
$θ=1.25\sinθ$