Is my answer wrong? I got (1/2)a^2(2π−sin(2Ø)−2cos(Ø))

[Kulla_9289]

I have the attached question. I got $\frac{1}{2}a^2(2π-sin(2Ø)-2cos(Ø))$. But the question asks me to write it in terms of $a^2$ and $Ø$ only. The process was lengthy but basically, I separated the circle into three parts and found the entire area for each of the sectors for each of the sectors; then, I subtracted the area of a triangle from the area of a sector; then, I added all the three subtracted areas of the shaded region to get to the answer above.

 

[blamocur]

Why do you believe that your answer is wrong? If you are given a different answer can you post it?
Also, I find your description of the solution confusing. Can you post intermediate formulas for for your steps?

 

[Kulla_9289]

Area of sector: 1/2(r^2)(θ)
Area of triangle: 1/2(a)(b)(sinC)

The θ is π - 2Ø

I think it's wrong because they only want it in terms of $a^2$ and $Ø$.

 

[topsquark]

Area of sector: 1/2(r^2)(θ)
Area of triangle: 1/2(a)(b)(sinC)

The θ is π - 2Ø

I think it's wrong because they only want it in terms of $a^2$ and $Ø$.

If you are asking if the $\pi$ is allowed to be there, then yes, it's okay. I mean, there's a 2 in there as well, but no one would object to that.

-Dan

 

[blamocur]

Area of sector: 1/2(r^2)(θ)

Which sector is this?

Area of triangle: 1/2(a)(b)(sinC)

What is 'b' ? What is 'C' ?

 

[Steven G]

1st, the problem does NOT ask for the area denoted in terms of a² and Ø. Where do you see that??
1/2a²(2π−sin(2Ø)−2cos(Ø)) = 1/2a²(2π−2sin(Ø)cos(Ø)−2cos(Ø)). Now this only has a and Ø.

 

[Kulla_9289]

So the answer is correct? I thought that sin and cos cannot be in the question.

 

[Dr.Peterson]

So the answer is correct? I thought that sin and cos cannot be in the question.

Yes.



The phrase "in terms of $a$ and $\phi$ only" means that those are the only variables in the expression.

 

[MaxMath]

I got $\frac{1}{2}a^2(2π-sin(2\phi)-2cos(\phi))$. But the question asks me to write it in terms of $a^2$ and $\phi$ only.

The problem requires the answer to be in terms of $a$ (not $a^2$) and $\phi$ only. But no matter which, your answer already satisfies this condition, doesn't it? That is, $a$ and $\phi$ are the only variables in your answer. (Though I have not confirmed your answer is indeed correct.)

 

[Kulla_9289]

Was my approach correct? Is this truly the correct answer?

 

[MaxMath]

Your process looks more complicated than necessary. It can simply calculated by subtracting the area of the sector from the area of the circle. The radius of the sector, and its angle, can be calculated.

 

[Kulla_9289]

How can you calculate the angle of the sector?

 

[MaxMath]

How can you calculate the angle of the sector?

It’s just double of $$\phi$$.

 

[Kulla_9289]

Why?

 

[Kulla_9289]

y is the radius of the sector. You cannot possibly get the answer is terms of a and ϕ only.

 

[MaxMath]

Why?

Draw a line between the centre of the circle and that of the sector. You can then find out.

 

[MaxMath]

y is the radius of the sector. You cannot possibly get the answer is terms of a and ϕ only.

You then need to find a way to express y in terms of the other two variables. This is not too difficult.

 

[MaxMath]

By the way, I think your answer is not correct. There will be a term with $\phi$ outside of sin or cos. Draw auxiliary lines, look for patterns, symmetries, broaden your territory bit by bit while keeping a big picture view.

 

[Dr.Peterson]

@MaxMath is correct: your answer, which I hadn't looked at closely, is wrong. Please show the details of your work.

This picture may help in finding y:

​

 

[Kulla_9289]

I have thought about it and used a circle theorem to get to $(a)^2(π-2ϕ)+\frac{1}{2}(a^2)(4ϕ)-\frac{1}{2}(y^2)(ϕ)$.
Edit: I have simplified it to $πa^2-\frac{1}{2}y^2ϕ$. How do I remove the y, as the question only wants it in terms of a and ϕ?