Circle Geometry

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October 7th 2008, 12:52 AM
xwrathbringerx

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Circle Geometry 2

Points A,B and C lie on a circle. The length of the chord AB is a constant k. Let angle ACB = a degrees and angle ABC = b degrees.

i) Why is a degrees always constant?

ii) How does the sum of the lengths of the chords AC and BC become (k/sin a)(sin b + sin(a + b))?

iii) When b = 90 - a/2, what is the expression for S?

Could someone please help me (especially with ii)?

Thanx very much
October 7th 2008, 05:39 PM
Nacho

i) because <a always takes the same arc AB

ii) using "sin theorem"

$<br /> \frac{{\sin a}}<br /> {K} = \frac{{\sin b}}<br /> {{AC}} \Leftrightarrow AC = \frac{{K\sin b}}<br /> {{\sin a}}<br />$

again

$<br /> \frac{{\sin a}}<br /> {K} = \frac{{\sin \left( {180 - (a + b)} \right)}}<br /> {{BC}} \Leftrightarrow \frac{{\sin a}}<br /> {K} = \frac{{\sin \left( {a + b} \right)}}<br /> {{BC}} \Leftrightarrow BC = \frac{{K\sin \left( {a + b} \right)}}<br /> {{\sin a}}<br />$

suming

$<br /> AC + BC = \frac{{K\sin b}}<br /> {{\sin a}} + \frac{{K\sin \left( {a + b} \right)}}<br /> {{\sin a}} = \frac{K}<br /> {{\sin a}}\left( {\sin b + \sin \left( {a + b} \right)} \right)<br />$

What is S?
October 7th 2008, 05:45 PM
xwrathbringerx

Oops sori

S = sum of the lengths of the chords AC and BC
October 7th 2008, 07:03 PM
Nacho

Quote:

Originally Posted by xwrathbringerx

Oops sori

S = sum of the lengths of the chords AC and BC

ok, then only resitute "b" in ii)
October 11th 2008, 03:21 AM
xwrathbringerx

What exactly do you get for iii

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