matt@oddjob.UChicago.UUCP (Matt Crawford) (07/09/84)
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.
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Matt University ARPA: crawford@anl-mcs.arpa
Crawford of Chicago UUCP: ihnp4!oddjob!matt