if A,B,C,D be angle of a cyclic quadrilateral taken in order then Cos A + cos b + cos C + cos D equal to
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value of Cos A + cos b + cos C + cos d is equal to zero
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Cyclic quadrilateral is a quadrilateral whose all vertices lie on a single circle.
In cyclic quadrilateral, the sum of opposite angles is 180°
i•e, A+C= 180° and B+D= 180°
cosA+cosC = 2cos(A+C/2)cos(A-C/2)
cosA+cosC = 2cos(180°/2)cos(A-C/2)
cosA+cosC= 0 {as cos90°=0}
cosB+cosD = 2cos(B+D/2)cos(B-D/2)
cosB+cosD = 2cos90°cos(B-D/2)
this is your answer....
it is helpful for you....
In cyclic quadrilateral, the sum of opposite angles is 180°
i•e, A+C= 180° and B+D= 180°
cosA+cosC = 2cos(A+C/2)cos(A-C/2)
cosA+cosC = 2cos(180°/2)cos(A-C/2)
cosA+cosC= 0 {as cos90°=0}
cosB+cosD = 2cos(B+D/2)cos(B-D/2)
cosB+cosD = 2cos90°cos(B-D/2)
this is your answer....
it is helpful for you....
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