CBSE Class 11-science Answered
Let A, B and C be three vertices of a triangle ABC. It is known that a unique circle can be drawn through any three points. Let O be the centre of such circle. Extend AO to a point D on the circle as shown below. Join CD. Let R be the radius of the circum-circle of triangle ABC i.e. OA = OD = R.
ACD is a triangle in a semicircle and therefore angle ACD = 90 degree.
In right angled ACD,
sin D = AC/AD
i.e. sin D = AC/2R (1)
As angle B = angle D (because they stand on same arc AC), (1) above may be written
sin B = AC/2R
so 2R = AC/sin B
i.e. 2R = b/sin B (2)
In a similar way, (drawing diameter from C etc) it can be shown that
2R = BC/sin A
i.e. 2R = a/sin A (3)
and
2R = AB/sin C
i.e. 2R = c/sin C (4)
From (2), (3) and (4);
a/sin A = b/sin B = c/sin C = 2R
Hence, proved