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 Floating Tangent (Posted on 2009-10-02)

 No Solution Yet Submitted by brianjn No Rating

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`In addition to the figure -let I be the enter of the circleand P the point of tangency of RTwith the circle.`
`If t = tan((angle PIC)/2), then it is easy to show that the area oftriangle RST is`
`   sqrt(3)*(3 - t^2)  -------------------       1 - 3*t^2`
`Note: Angle PIC can be measured CW       or CCW. Thus, we can have one      or two positions for point P.`
`1a)   sqrt(3)*(3 - t^2)     ------------------- = 3*sqrt(3).           1 - 3*t^2`
`     or t = 0.`
`     Thus, P = C.`
`1b)   sqrt(3)*(3 - t^2)     ------------------- = 8.           1 - 3*t^2`
`                   8 - 3*sqrt(3)     or t = sqrt[ --------------- ].                    24 - sqrt(3)`
`     Thus, angle PIC =`
`                      8 - 3*sqrt(3)     2*arctan{ sqrt[ --------------- ] }                       24 - sqrt(3)`
`     ~= 39.073817 degrees.`
`1c)   sqrt(3)*(3 - t^2)     ------------------- = 6*sqrt(3)           1 - 3*t^2`
`     or t = sqrt[ 3/17 ].`
`     Thus, angle PIC = `
`     2*arctan{ sqrt[ 3/17 ] }`
`     ~= 45.572996 degrees.   `
`2)   For angle RTS to be a right angle,     angle PIC must equal 30 degrees.`
`     Thus, area [RST] `
`        sqrt(3)*(3 - tan(15)^2)     = -------------------------            1 - 3*tan(15)^2`
`     = 2*sqrt(3) + 3`
`     ~= 6.464102 `
` `

 Posted by Bractals on 2009-10-03 14:51:58

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