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In A B C ,C=60^0a n dB=45^0dot Line joi...

In ` A B C ,C=60^0a n dB=45^0dot` Line joining vertex A of triangle and its circumcenter `(O)` meets the side `B CinD` Find the ratio `B D : D C` Find the ratio `A O : O D`

Text Solution

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In the figure, `angle AOC = 2B`
`rArr angle DOC = pi - 2B`
Similarly, `angleBOD = pi - 2C`
Now, `(BD)/(CD) = ((1)/(2) BD.OP)/((1)/(2) CD.OP)`
`= ("Area of " Delta BOD)/("Area of " Delta COD)`
`= ((1)/(2) OB.OD sin angleBOD)/((1)/(2) OC.OD sin angleCOD)`
`= (sin (pi -2C))/(sin (pi-2B)) = (sin 2C)/(sin 2B)`
`= (sin 120^(@))/(sin 90^(@))`
`= (sqrt3)/(2)`
Now in `Delta OPB, OP = OB cos A = R cos A`
Also `anglePOD = anglePOC - angleDOC`
`= A - (pi - 2B)`
`= pi - B - C - pi + 2B`
`= B - C`
Now in `DeltaOPD, OD = (PO)/(cos(B -C)) = (R cos A)/(cos (B -C))`
`rArr (AO)/(OD) = (R)/((R cos A)/(cos (B -C)))`
`= (cos(B-C))/(cosA) = (cos 15^(@))/(cos 75^(@)) = (cos 15^(@))/(sin 15^(@)) = 2 sqrt3`
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