A certain triangle ABC has D on AB and E on AC such that AD=DB=BC=CE and AE=ED. What is the measure of angle BAC?
(In reply to
solution by Charlie)
Going to http://www.1728.com/cubic2.htm to see how to solve cubic equations, I found the value for x (one side of the small isosceles triangle ADE), to be
((37+9*sqrt(17))^(1/3) + (37-9*sqrt(17))^(1/3) - 2) / 3
The angle is still arccos(1/(2*x)) or
arcsec(2*((37+9*sqrt(17))^(1/3) + (37-9*sqrt(17))^(1/3) - 2) / 3)
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Posted by Charlie
on 2005-10-24 15:18:40 |
re: solution -- closed form
