# Radians

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Thread starter 8 years ago
#1
The diagram shows a circle with centre O and radius 6cm. The chord PQ divides the circle into a minor segement R1 of area A1cm^2 and a major segment R2 of area A2cm^2. The chord PQ subtends an angle theta raians at O.

Given that A2 = 3A1 and f(theta) = 2theta - 2 sin theta - pi

a) Prove that f(theta) = 0

b)
evaluate f(2.3) and f(2.32) and deduce that 2.3<theta<2.3.

So i did the area of the minor sector minus the area of the circle and this is equal to 3 x the minor segment

i have:

pi r^2 - 18theta = 3 [18(theta - sin theta) ]

Is this correct?
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Thread starter 8 years ago
#2
Anyone?
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8 years ago
#3
K ill try

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8 years ago
#4
(Original post by Tynos)
The diagram shows a circle with centre O and radius 6cm. The chord PQ divides the circle into a minor segement R1 of area A1cm^2 and a major segment R2 of area A2cm^2. The chord PQ subtends an angle theta raians at O.

Given that A2 = 3A1 and f(theta) = 2theta - 2 sin theta - pi

a) Prove that f(theta) = 0

b)
evaluate f(2.3) and f(2.32) and deduce that 2.3<theta<2.3.

So i did the area of the minor sector minus the area of the circle and this is equal to 3 x the minor segment

i have:

pi r^2 - 18theta = 3 [18(theta - sin theta) ]

Is this correct?
How did you set this equation up?

I started from (Area major segment) + (Area minor segment) = Area of circle = 36pi.

Also:
Area major segment = 3 x (Area minor segment) (given)
and
(Area minor segment) = (Area of sector OPQ) - (Area of triangle OPQ)
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8 years ago
#5
I think it's right. It's kinda hard. Where does functions come in c2?
I'm doing radians at school. At alternate method is to directly find the A1 and then use the equation.Here.

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Thread starter 8 years ago
#6
(Original post by StUdEnTIGCSE)
I think it's right. It's kinda hard. Where does functions come in c2?
I'm doing radians at school. At alternate method is to directly find the A1 and then use the equation.Here.

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Thanks but where did you get the 36 pi?
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8 years ago
#7
(Original post by Tynos)
Thanks but where did you get the 36 pi?
36pi is the area of the circle - you're told that the radius is 6cm.
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5 years ago
#8
(Original post by davros)
36pi is the area of the circle - you're told that the radius is 6cm.
how do you know that you need to use the area of the circle to work out the major segment because the major segment would simply be 3 times bigger than the minor segment isn't it? Thanks I know it is probably really late
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5 years ago
#9
(Original post by coconut64)
how do you know that you need to use the area of the circle to work out the major segment because the major segment would simply be 3 times bigger than the minor segment isn't it? Thanks I know it is probably really late
If the major segment is 3x the area of the minor segment then the total area of both (i.e. the area of the circle) is 3x minor segment area + 1x minor segment area = 4x minor segment area

Therefore minor segment area = 1/4 of circle
So the major segment area = 3/4 area of circle ---> you can use area of circle to find area of major segment.
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5 years ago
#10
(Original post by Pronged Lily)
If the major segment is 3x the area of the minor segment then the total area of both (i.e. the area of the circle) is 3x minor segment area + 1x minor segment area = 4x minor segment area

Therefore minor segment area = 1/4 of circle
So the major segment area = 3/4 area of circle ---> you can use area of circle to find area of major segment.
So does it mean that a circle is made up of 4 minor segments? Or just this circle and if it is optional then how do people know they need to do this in this question because I didn't think of doing that
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5 years ago
#11
(Original post by coconut64)
So does it mean that a circle is made up of 4 minor segments? Or just this circle and if it is optional then how do people know they need to do this in this question because I didn't think of doing that

When you split a circle into 2 parts using a chord, there is a smaller part (minor segment) and a larger part (major segment).
The two added together make up a whole circle. In this particular circle...

"The chord PQ divides the circle into a minor segement R1 of area A1cm^2 and a major segment R2 of area A2cm^2.Given that A2 = 3A1 and f(theta) = 2theta - 2 sin theta - pi..."... the area of the major segment is 3 times the area of the minor segment.
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5 years ago
#12
(Original post by Pronged Lily)
When you split a circle into 2 parts using a chord, there is a smaller part (minor segment) and a larger part (major segment).
The two added together make up a whole circle. In this particular circle...

"The chord PQ divides the circle into a minor segement R1 of area A1cm^2 and a major segment R2 of area A2cm^2.Given that A2 = 3A1 and f(theta) = 2theta - 2 sin theta - pi..."... the area of the major segment is 3 times the area of the minor segment.
Oh I think I get it now because they add up to make a circle. Thanks
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