1.

If A,B,C and D are angles of a quadrilateral and sinA/2sinB/2sinC/2sinD/2=1/4,prove that A=B=C=D=pi//2

Answer»

Solution :`(2sinA/2sinB/2)(2sinC/2sinD/2)=`
`rArr {cos(A-B)/2 - cos(A+B)/2}{cos(C-D)/2-cos(C+D)/2}=1`
Since, `A+B=2pi-(C+D)`, the above EQUATION becomes,
`rArr {cos(A-B)/2 -cos(A+B)/2}{cos(C-D)/2 + cos(A+B)/2}=1`
`rArr cos^(2)(A+B)/2 -cos(A+B)/2{cos(A-B)/2-cos(C-D)2}+1-cos(A-B)/2cos(C-D)/2=0`
This is a QUADRATIC equation in `cos(A+B)/2` which has real ROOTS.
`rArr {cos(A-B)/2-cos(C-D)/2}^(2)-4{1-cos(A-B)/2.cos(C-D)} ge 0`
`(cos(A-B)/2 +cos(C-D)/2)^(2) ge 4`
`rArr cos(A-B)/2 + cos(C-D)/2 ge`, Now both `cos(A-B)/2` and `cos(C-D)/2 le 1`
`rArr cos(A-B)/2=1` and `cos(C-D)/2=1`
`rArr (A-B)/2 =0 = (C-D)/2`
`A=B, C=D`
Similarly, A=C,B=D `rArr A=B=C=D=pi/2`


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