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From: matt@oddjob.UChicago.UUCP (Matt Crawford)
Newsgroups: net.math,net.puzzle
Subject: Re: Geometry Puzzle    SOLUTION
Message-ID: <326@oddjob.UChicago.UUCP>
Date: Mon, 9-Jul-84 16:18:52 EDT
Article-I.D.: oddjob.326
Posted: Mon Jul  9 16:18:52 1984
Date-Received: Tue, 10-Jul-84 02:29:12 EDT
References: <2861@ecsvax.UUCP>
Organization: U. Chicago: Astronomy & Astrophysics
Lines: 29

Given information:
	BAC = 40,  ABE = ADE = 10,  AB = AD,  B != D.

(1) Exterior angle thm ==> AEB = BAC - ABE = 40 - 10 = 30

(2) sin AEB   sin ABE
    ------- = -------    (law of sines)
      AB        AE

              sin ADE
            = -------    (given ABE=ADE)
                AE

              sin AED
            = -------    (law of sines)
                AD

              sin AED
            = -------    (given AD=AB)
                AB

    Therefore  sin AEB = sin AED.  But  B != D  ==>  AEB != AED,
    so  AED = 180 - AEB = 180 - 30 = 150

(3) DAE = 180 - ADE - AED = 180 - 10 - 150 = 20.

___________________________________________________________
Matt		University	ARPA: crawford@anl-mcs.arpa
Crawford	of Chicago	UUCP: ihnp4!oddjob!matt