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In triangleABC, angleB=60^(@) and angleC...

In `triangleABC, angleB=60^(@)` and `angleC = 75^(@)`. If D is a point on side BC such that `("ar"(triangleABD))/("ar"(triangleACD))=1/sqrt(3)`
find `angleBAD`

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`angle B = 60^(@), angleC=75^(@)`
`therefore angleA=180^(@) - (angleB+angleC) = 180^(@) - (60^(@) + 75^(@))= 45^(@)`
Let `angleDAC = theta`, then `angleBAD = 45^(@)-theta`
`therefore ("ar"(triangleBAD))/("ar"(triangleADC)) = (1/2 CAD.sin(45-theta))/(1/2b.AD sintheta)`
`rArr 1/sqrt(3) = (c sin(45^(@)-theta))/(b sintheta)`

`rArr 1/sqrt(3) (sinC(sin45^(@). costheta-cos45^(@).sintheta)/(sinB. sintheta))`
`rArr (sqrt(3)+1)/(2) (costheta - sintheta)=sin theta rArr cos theta = sqrt(3) sintheta`
`rArr tantheta = 1/sqrt(3)`
`rArr theta=30^(@)`
`therefore angleBAD = 45^(@)-theta= 45^(2)-30^(@)=15^(@)`
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