hi guys i need help on this question

Prove that, for any triangle ABC inscribed in a circle of radius r,

a over sinA= b over sinB= c over sinC= 2r.

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- September 20th 2006, 04:27 PMmathishardthe proof of law of sines
hi guys i need help on this question

Prove that, for any triangle ABC inscribed in a circle of radius r,

a over sinA= b over sinB= c over sinC= 2r. - September 20th 2006, 05:46 PMThePerfectHacker
Ah! The elegant extended law of sines.

Below is a hint how to solve problem. I drew a line from a vertex through the diameter (which is not one of the vertices) until it intersected the circumfurence, in blue. THen I extended the red line. Note the two angles which are marked are equal because they create equal arcs.

The rest I leave to thee. - September 20th 2006, 07:10 PMmathishard
i know how to draw the triangle and stuff already it just that i don't know how to proof it.

- September 20th 2006, 08:05 PMThePerfectHacker
1)Since DC is the diameter we know that <DAC is right.

2)Also, B=D.

3)Thus, sin B= sin D

4)But, sin D=AC/DC

5)Thus, sin B=AC/DC

6)Thus, 1/sin B=DC/AC

7)Thus, AC/sin B=DC

8)But, DC=2r

9)Thus, AC/sin B=2r

10)Similarly, AB/sin C=2r and BC/sin A=2r

11)Thus, BC/sin A=AC/sin B=AB/sin C=2r

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This shows that side divided by opposite angle is 2r.