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.
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.
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.