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Triangle In Circles GCSE

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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?
Reply 1
Avatar for hp4300
hp4300
12
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
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)
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!!!
Never mind it’s sine rule, just wondered why you used cosine but I got the same answer 👍🏻
Reply 5
Avatar for hp4300
hp4300
12
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
Original post by Angel_Chen
You are not allowed to give step by step solutions, only hints!!!


Why can’t he?
Original post by Y11_Maths
Why can’t he?


IDK it's a rule in the maths forum
Reply 8
Avatar for hp4300
hp4300
12
She*
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?
https://www.thestudentroom.co.uk/showthread.php?t=4919248

answering guidlines
Sine rule*
Reply 12
Avatar for hp4300
hp4300
12
Original post by Angel_Chen


Sorry, I'm new to TSR should I remove my post?

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