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if A,B,C,D are the angles of a cyclic quadrilateral, the p.t. i)cosA+cosB+cosC+cosD=0 ii)sinA-sinC=sinD-sinB. |
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Answer» -STEP explanation:SINCE the quadrilateral ABCD is CYCLIC, we haveA+C=180o ;B+D=180o HENCE, cosA=cos(180o −C)=−cosC...(1) and cosB=cos(180o −C)=−cosD...(2)Adding (1) and (2) we getcosA+cosB=−cosC−cosDor cosA+cosB+cosC+cosD=0 |
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