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Path: utzoo!watmath!ljdickey
From: ljdickey@watmath.UUCP
Newsgroups: net.math
Subject: Re And another high-school problem
Message-ID: <5917@watmath.UUCP>
Date: Thu, 6-Oct-83 20:38:48 EDT
Article-I.D.: watmath.5917
Posted: Thu Oct  6 20:38:48 1983
Date-Received: Fri, 7-Oct-83 11:09:27 EDT
Sender: ljdickey@watmath.UUCP
Organization: U of Waterloo, Ontario
Lines: 20

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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