Ans 1:√3 BD=2r this gives BC=2rsin(@) &CD= 2rcos(@) where angleODC=angleADE=@ area of rectangle=4r^2[sin@cos@] =2r^2sin2@ πr^2/2r^2sin2@=π/√3 sin2@=√3/2 this gives @=30 degree Tan@=AE/DE=1/√3
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Ans 1:√3
BD=2r this gives BC=2rsin(@) &CD= 2rcos(@) where angleODC=angleADE=@
area of rectangle=4r^2[sin@cos@]
=2r^2sin2@
πr^2/2r^2sin2@=π/√3
sin2@=√3/2 this gives @=30 degree
Tan@=AE/DE=1/√3
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Mja kar diya sir aapne itne important qus helpful h bahut achhe concept h… Sir isi tarah continue… Rakhna sir aap bahut achhha karya kar rhe h ham students k liye… A lot of thanks sir ji for this support….. Magician of maths…….. Great personality…….
Your questions' collection for every video in the ongoing series is awesome…
Sir…. Plz tell us ki anuj sengar sir ki class q nhi aa rhi h???? This is not fair ki pathshala ab koi msg nhi deta hai.. eagaerly waiting for his class.. nd completely dependent on u all.. Plz reply
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