In a triangle ABC,the value of sinA+sinB+sinC is :
4sinA2sinB2sinC2
4cosA2cosB2cosC2
4cosA2sinB2sinC2
4cosA2sinB2cosC2
Finding the value of sinA+sinB+sinC in a triangle.
In any triangle, the sum of all the interior angles is always 180°.
Therefore, In the triangle ABC, A+B+C=180°
Now, solving for sinA+sinB+sinC:
sinA+sinB+sinC=2sinA+B2cosA-B2+sinC=2sinA+B2cosA-B2+2sinC2cosC2=2sin180°-C2cosA-B2+2sinC2cosC2=2cosC2cosA-B2+2sinC2cosC2=2cosC2cosA-B2+sinC2=2cosC2cosA-B2+sin180°-(A+B)2=2cosC2cosA-B2+cosA+B2=4cosC2cosB2cosA2sinA+sinB+sinC=4cosA2cosB2cosC2
Hence, Option (B) is correct.
In △ABC,R=10, then the value of r(a+b+c)abc is