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

K

Kulla_9289

Junior Member
Joined
Apr 18, 2022
Messages
214
I have the attached question. I got [imath]\frac{1}{2}a^2(2π-sin(2Ø)-2cos(Ø))[/imath]. But the question asks me to write it in terms of [imath]a^2[/imath] and [imath]Ø[/imath] 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.
 

Attachments

  • 1680276259305.png
    1680276259305.png
    43 KB · Views: 19
Last edited:
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?
 
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 [imath]a^2[/imath] and [imath]Ø[/imath].
 
Kulla_9289 said:
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 [imath]a^2[/imath] and [imath]Ø[/imath].
Click to expand...
If you are asking if the [imath]\pi[/imath] 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
 
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.
 
So the answer is correct? I thought that sin and cos cannot be in the question.
 
Kulla_9289 said:
I got [imath]\frac{1}{2}a^2(2π-sin(2\phi)-2cos(\phi))[/imath]. But the question asks me to write it in terms of [imath]a^2[/imath] and [imath]\phi[/imath] only.
Click to expand...
The problem requires the answer to be in terms of [imath]a[/imath] (not [imath]a^2[/imath]) and [imath]\phi[/imath] only. But no matter which, your answer already satisfies this condition, doesn't it? That is, [imath]a[/imath] and [imath]\phi[/imath] are the only variables in your answer. (Though I have not confirmed your answer is indeed correct.)
 
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.
 
y is the radius of the sector. You cannot possibly get the answer is terms of a and ϕ only.
 
Kulla_9289 said:
y is the radius of the sector. You cannot possibly get the answer is terms of a and ϕ only.
Click to expand...
You then need to find a way to express y in terms of the other two variables. This is not too difficult.
 
By the way, I think your answer is not correct. There will be a term with [imath]\phi[/imath] outside of sin or cos. Draw auxiliary lines, look for patterns, symmetries, broaden your territory bit by bit while keeping a big picture view.
 
@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:

1680358502626.png​
 
I have thought about it and used a circle theorem to get to [imath](a)^2(π-2ϕ)+\frac{1}{2}(a^2)(4ϕ)-\frac{1}{2}(y^2)(ϕ)[/imath].
Edit: I have simplified it to [imath]πa^2-\frac{1}{2}y^2ϕ[/imath]. How do I remove the y, as the question only wants it in terms of a and ϕ?
 
Last edited:
You must log in or register to reply here.
Top