Y1_UniMaths
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I jotted down what I knew but I can’t think of where to start on this problem. Any ideas/starting points?
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hp4300
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#2
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Let M be the mid point of line AB:

Line CM is radius of circle 1 + 2
So CM is 5+4
CM = 9

Where tangent AC touches the circumference of the circle is point P
Join line CM to P to make a triangle.
Remember that PM is the radius of circle 1 so it's 5 cm long.
Sin-1(5/9) = 33.75 degrees
Angle ACM is 33.76 degrees

Create another triangle MCB
Rember that angle ACM = MCB
BC = 9/ cos(33.75)
BC = 10.82

Angle ACB = 33.75 x 2
= 67.5
Total area = 1/2 ab sinC
1/2 X 10.82 x 10.82 x sin(67.5)
= 54.1 cm^2
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Y1_UniMaths
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#3
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(Original post by hp4300)
Let M be the mid point of line AB:

Line CM is radius of circle 1 + 2
So CM is 5+4
CM = 9

Where tangent AC touches the circumference of the circle is point P
Join line CM to P to make a triangle.
Remember that PM is the radius of circle 1 so it's 5 cm long.
Sin-1(5/9) = 33.75 degrees
Angle ACM is 33.76 degrees

Create another triangle MCB
Rember that angle ACM = MCB
BC = 9/ cos(33.75)
BC = 10.82

Angle ACB = 33.75 x 2
= 67.5
Total area = 1/2 ab sinC
1/2 X 10.82 x 10.82 x sin(67.5)
= 54.1 cm^2
Why is BC 9/ cos(33.75)
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Angel_Chen
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(Original post by hp4300)
Let M be the mid point of line AB:

Line CM is radius of circle 1 + 2
So CM is 5+4
CM = 9

Where tangent AC touches the circumference of the circle is point P
Join line CM to P to make a triangle.
Remember that PM is the radius of circle 1 so it's 5 cm long.
Sin-1(5/9) = 33.75 degrees
Angle ACM is 33.76 degrees

Create another triangle MCB
Rember that angle ACM = MCB
BC = 9/ cos(33.75)
BC = 10.82

Angle ACB = 33.75 x 2
= 67.5
Total area = 1/2 ab sinC
1/2 X 10.82 x 10.82 x sin(67.5)
= 54.1 cm^2
You are not allowed to give step by step solutions, only hints!!!
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Y1_UniMaths
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#5
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Never mind it’s sine rule, just wondered why you used cosine but I got the same answer 👍🏻
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hp4300
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Sorry there is a typo I meant to say ACM is 33.75 degrees

Triangle MCB is a right angled triangle to find BC we use cos
Cos = A/H
Cos(33.75) = 9/H
9/ cos(33.75) = H
10.82 = H
10.82 = BC
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Y1_UniMaths
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#7
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(Original post by Angel_Chen)
You are not allowed to give step by step solutions, only hints!!!
Why can’t he?
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Angel_Chen
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(Original post by Y11_Maths)
Why can’t he?
IDK it's a rule in the maths forum
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hp4300
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#9
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#9
She*
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Y1_UniMaths
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#10
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(Original post by hp4300)
Sorry there is a typo I meant to say ACM is 33.75 degrees

Triangle MCB is a right angled triangle to find BC we use cos
Cos = A/H
Cos(33.75) = 9/H
9/ cos(33.75) = H
10.82 = H
10.82 = BC
They both work don’t they? Sohcahtoa or some rule to find BC?
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Angel_Chen
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#11
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https://www.thestudentroom.co.uk/sho....php?t=4919248

answering guidlines
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Y1_UniMaths
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#12
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#12
Sine rule*
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hp4300
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#13
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Sorry, I'm new to TSR should I remove my post?
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