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=90^{0} [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=90**^{0} , sin A = sin 90^{0} = 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.

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

Hence, for every condition we have, Similarly, we can prove, Combining these we get,

, which is the Sine Law in terms of circum-radius.

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