The Sine Law in terms of Circum-Radius
Let ABC be the triangle with all three vertex on the circumference of the circle with center O as shown in fig above.BO is drawn if the angle is either acute or obtuse as shown in fig(i) and fig(ii). Also, BO is produced to D to form the diameter BD as shown in fig. [C coincides D in fig (iii)]Then, circum-radius=OB=R(radius of circle) and BD=2R and angle BCD=900 [in fig (i) and (ii)] being angle at semi-circle. To Prove:
1. For a triange as shown in fig(i)
BAC=BDC [angle subtended by same segment]
i.e., angle A = angle D.........(i)In triangle BDC,
2. For a triangle as shown in fig(ii)
In triangle BDC,
3. For a triangle as shown in fig(iii)
BC = 2R = 2R sin A [A=900 , sin A = sin 900 = 1]
Hence, for every condition we have, Similarly, we can prove, Combining these we get,
, which is the Sine Law in terms of circum-radius.
Please Log In to ask your question.