If abcd is a cyclic quadrilateral, then show that cos a + cos b + cos c + cos d = 0.
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4
a=180-c
so cosa=cos(180-c)=-cosa
similarily cosb=-cosd
so we get 0
so cosa=cos(180-c)=-cosa
similarily cosb=-cosd
so we get 0
Answered by
6
A+C = 180. & C=180-A
B+D = 180. & D=180-B
cosA + cosB + cosC + cosD = 0
LHS :
=cosA + cosB + cos(180-A) + cos(180-B)
=cosA + cosB - cosA - cosB
=0
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