Triangle ABC has a point D on side BC such that BA=AD=DC=1. What is angle ABD when the area of the triangle is maximized, and what is the maximum area?
Since AD = AB, angle ADB = angle ABD.
Let t = angle ABD. Therefore, 0 < t < 90 and BD = 2*cos(t).
BA*BC*sin(t) 1*(2*cos(t) + 1)*sin(t)
Area(ABC) = -------------- = -------------------------
2 2
(4*cos(t)^2 + cos(t) -2) sqrt(33) - 1
Area(ABC)' = -------------------------- = 0 ==> cos(t) = --------------
2 8
-sin(t)*(8*cos(t) + 1)
Area(ABC)" = ------------------------ < 0 for 0 < t < 90
2
Therefore,
Angle ABD ~= 53.624808 degrees for maximum
Area(ABC) ~= 0.880086
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Posted by Bractals
on 2005-09-19 16:22:44 |
Solution
