1.

P is a point inside the equilateral traingle ABC such that PA=1cm, PB=sqrt(2)cm and PC=sqrt(3).Then the side of the traingle("in cm") equal.

Answer»

`SQRT(6-2sqrt(3))`
`sqrt(6-sqrt(3))`
`sqrt(3+sqrt(6))`
`sqrt(3+2sqrt(6))`

Solution :In`DeltaPAB`
`cos theta=(a^(2)+1-2)/(2A)=(a^(2)-1)/(2a)`
In `DeltaPAC`
`cos (60-theta)=(1+a^(2)-3)/(2a)=(a^(2)-2)/(2a)`
`therefore cos theta +cos (60-theta)=(2a^(2)-3)/(2a)implies cos(30-theta)=(2a^(2)-3)/(2sqrt(3a))`
`"ALSO", ""costheta-cos(60-theta)=(1)/(2a)impliessin(30-theta)=(1)/(2a)`
`implies((2a^(2)-3)/(2sqrt(3a)))^(2)+((1)/(2a))^(2)=1`
`implies4a^(4)-24a^(2)+12=0`
`impliesa^(2)=3+sqrt(6)` or `a=sqrt(3+sqrt(6))`


Discussion

No Comment Found

Related InterviewSolutions