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At the endpoint and midpoint of a circul...

At the endpoint and midpoint of a circular are AB, tangent lines are drawn, and the points, A and B are jointed with a chord. Prove that the ratio of the areas of the triangles thus formed tends to 4 as the arc AB decreases infinitely.

Text Solution

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Let `Delta_(1)="Area of triangle "ABC`
and `Delta_(2)="Area of triangle "CDE`
In `AFO,AF=Rsintheta" and "OF=Rcostheta`
`MF=OM-OF=R-Rcostheta`
In triangle `OAC, OC=Rsectheta`
`:." "CM=OC-OM=Rsectheta-R`
`:." "CF=CM+MF=Rsectheta-Rcostheta`
So, `Delta_(1)=(1)/(2)AB.CF`
`=(1)/(2)2Rsintheta.R(sectheta-costheta)`
`=R^(2)tantheta(1-cos^(2)theta)`
In triangle `CMD,DM=CMcot theta`
`=(Rsectheta=-R)cot theta`
`:." "Delta_(2)=(1)/(2)2(Rsectheta-R)cottheta.(Rsectheta-R)`
`=R^(2)((1-cos^(2)theta)^(2))/(cos^(2)theta.tantheta)`
`:." "(Delta_(1))/(Delta_(2))=(tantheta(1-cos^(2)theta)cos^(2)thetatantheta)/((1-costheta)^(2))`
`=(sin^(2)theta(1+costheta))/((1-costheta))`
`:." "underset(thetato0)lim(Delta_(1))/(Delta_(2))=underset(thetato0)lim(sin^(2)theta(1+costheta))/((1-costheta))`
`=underset(thetato0)lim(sin^(2)theta(1+costheta)^(2))/(1-cos^(2)theta)`
`=underset(thetato0)lim(1+costheta)^(2)=4`
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