ljdickey@watmath.UUCP (10/07/83)
About the question of the isosceles triangle with angles 80, 80, and 20.
The question amounts to this: Bisect one of the long sides (legs)
and then find all the angles.
If the original triangle is ABC and <BAC = 20, and D is the bisector
of side AB, label the angle at C, made with the base, by ALPHA = <BCD.
The quickest solution (takes less than 5 minutes to think of) uses the
law of sines on the triangle ACD. We have:
SIN 20 SIN (80-ALPHA) SIN (80+ALPHA)
----------- = -------------------- = ------------------
X T 2 T
and here T is half of the length of the side AB. Some easy trigonometry
gives the equation TAN ALPHA = ( TAN 80 ) / 3.
My favourite machine tells me that ALPHA is 62.1220128556668995.
What do you get?
--
Lee Dickey, University of Waterloo. (ljdickey@watmath.UUCP)
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