a/sinA=b/sinb=c/sinC= 2r

Question

In any triangle ABC show that a/sinA=b/sinb=c/sinC=2R

Answer 1

Draw a circle centre X and mark three points A, B and C on the circumference. Join the points to make triangle ABC. Draw radii AX, BX and CX=r. The triangles AXB, BXC and AXC are isosceles. Angle AXC=2ABC, angle BXC=2BAC and angle AXB=2ACB (angle at the centre of a circle is twice the angle at the circumference). The equal angles of the isosceles triangles can be calculated: BAX=ABX=(1/2)(180-AXB)=90-ACB, and similarly the other angles.

Applying the sine rule to just one isosceles triangle we have AB/sin2ACB=XB/sinBAX=r/cosACB. But sin2ACB=2sinACBcosABC, so r/cosACB=c/2sinACBcosACB. Therefore, r=c/2sinACB and 2r=c/sinC=a/sinA=b/sinB.

 

Answer 2

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