Hard question, Need Help!!! Watch

Tiri
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The chord AB of a circle divides the circle into two portions whose areas are in the ratio 3:1. If AB makes an angle θ with the diameter passing through A, show that θ satisfies the equation;


Sin(2θ)= π/2 -2θ


I ended up getting Sin (
θ) = θ - π/2 , from the minor segment ratio but I dont know where to go from here...
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ghostwalker
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(Original post by Tiri)
The chord AB of a circle divides the circle into two portions whose areas are in the ratio 3:1. If AB makes an angle θ with the diameter passing through A, show that θ satisfies the equation;


Sin(2θ)= π/2 -2θ


I ended up getting Sin (
θ) = θ - π/2 , from the minor segment ratio but I dont know where to go from here...
I suggest posting your working, as that can't be correct. Your equation has no solution in the interval 0 to pi/2, which is the valid range of theta.
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Tiri
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As you can see i posted a picture of the question, its from 1966


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ghostwalker
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(Original post by Tiri)

As you can see i posted a picture of the question, its from 1966
Very nice, but I can't see your working. Can't tell where you've gone wrong without that.
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TenOfThem
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(Original post by Tiri)
The chord AB of a circle divides the circle into two portions whose areas are in the ratio 3:1. If AB makes an angle θ with the diameter passing through A, show that θ satisfies the equation;


Sin(2θ)= π/2 -2θ[COLOR=#252525][FONT=Nimbus Roman No9 L]

Nice question
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Malgorithm
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I agree, had a lot of fun solving this one and watching everything cancel down really nicely. It's a shame these types of questions rarely appear in any of the exam papers.
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TenOfThem
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(Original post by Malgorithm)
It's a shame these types of questions rarely appear in any of the exam papers.
My thoughts exactly
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Tiri
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Are you lot all pretending that you actually done the question or are you gonna show how you did it? Something doesnt look right


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Malgorithm
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(Original post by Tiri)
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Are you lot all pretending that you actually done the question or are you gonna show how you did it? Something doesnt look right


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The question says that the angle theta is between the diameter and AB. You seem to have labelled it as the angle between OA and OB.

If you just fix that, your method seems to be just fine, it's just the values need altering a bit because you chose the wrong angle for theta.
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Tiri
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Name:  ImageUploadedByStudent Room1417366298.385846.jpg
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Size:  128.1 KB am i being an idiot or is there still something ive missed?


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TenOfThem
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(Original post by Tiri)
Name:  ImageUploadedByStudent Room1417366298.385846.jpg
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Size:  128.1 KB am i being an idiot or is there still something ive missed?


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The angle between AB and the diameter is theta
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Tiri
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LOL you both are telling me differnt things... How about someone posts solutions if youve done it.. I seem to be stuck on the last little, I cant prove its pie/2 minus theta, only pie minus 2theta.
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TenOfThem
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(Original post by Tiri)
LOL you both are telling me differnt things... How about someone posts solutions if youve done it.. I seem to be stuck on the last little, I cant prove its pie/2 minus theta, only pie minus 2theta.
No

We have both told you the same thing

The question states that the angle theta is between AB and the Diameter at A


You are not stuck on the last bit - you have not yet drawn the correct diagram so you do not have the correct angle at the centre
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Tiri
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done it..


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TenOfThem
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(Original post by Tiri)
done it..
good
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