A,B,C,D are the angles of a cyclic quadrilateral prove that sinA+sinB=sinC+sind
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A+C=180°
A=180-C
sinA=sin (180-C)
sinA=sinC ..........(1)
B+D=180°
B=180-D
sin B= sin(180-D)
sinB= sinD ............(2)
From (1) and (2)
sinA+sinB=sinC+sinD
A=180-C
sinA=sin (180-C)
sinA=sinC ..........(1)
B+D=180°
B=180-D
sin B= sin(180-D)
sinB= sinD ............(2)
From (1) and (2)
sinA+sinB=sinC+sinD
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