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<=
/p>
Calling the radius o=
f the
given circle R and the three s=
ides
of the triangle 2a , 2b , and 2c and applying Ptolomey to the
quadrilateral =
OMaCMb where Ma and Mb are the
midpoints of the original sides.
It is obvious <=
/o:p>
c =3D 2 ( a * SQRT (=
R^2
–b^2) + b * SQRT (R^2 –a^2) )/ R=
=
by substitution R=3D5 ; b =3D 5/2 ; a =3D 6/2 we =
get c =3D 91.92
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