Circle question (triangles too)

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SUMMARY

The discussion focuses on deriving the length of chord AB in a circle with center O and radius 6 cm, given the angle AOB as θ radians. The conclusion is that the length AB can be expressed as √{72(1 - cos θ)}. The cosine rule is applied to triangle AOB to establish this relationship, confirming that the radius squared (6 cm) is integral to the calculation. The solution emphasizes the importance of understanding the properties of chords in circles.

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  • Understanding of the cosine rule in trigonometry
  • Knowledge of circle geometry, specifically chords and sectors
  • Familiarity with radians and their application in angular measurements
  • Basic algebra for rearranging and factorizing equations
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  • Study the properties of chords in circles and their relationships with angles
  • Learn how to apply the cosine rule in various geometric contexts
  • Explore the derivation of arc lengths and sector areas in circle geometry
  • Investigate advanced trigonometric identities and their applications in geometry
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Homework Statement



O is the centre of the circle of radius 6cm, and A and B are two points on the circumference such that angle AOB = \theta radians.

Show that the length AB is equal to \sqrt {72 (1 - cos \theta)}

Homework Equations



I think the following might be relevant:

Cosine rule: \theta = cos^-1\frac{b^2 + c^2 - a^2}{2bc}

Area of sector = \frac{1}{2}r^2\theta

The Attempt at a Solution



I don't really know where to start here. I think pythagoras is involved somewhere. The radius 6 must therefore be squared, multplied by 2 = 72. But I don't want to work backwards from the given soultion.

Am I missing some theory about chords of circles?
 
Last edited:
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No matter, I solved it.

Treat AOB as a triangle. Apply cosine rule to find AB, rearrange and factorise under the squareroot.
 

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