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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. |
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Answer» `SQRT(6-2sqrt(3))` `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))`
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