Value of Angle DBCDate: 1/23/96 at 21:30:58 From: aviel Subject: I have this very annoying puzzle in front of me and I believe that the solution lies somewhere in the Internet but I can't find it. Can you help me? Imagine a triangle ABC on which angles B and C are equal to 80. D stands somewhere on AC in such a way that AD = BC. The question is to find the value of angle DBC. Aviel Date: 9/1/96 at 18:39:11 From: Doctor Jerry Subject: Re: I enjoyed working on this problem. My solution depends upon finding two equations involving BC and then eliminating this unknown. Use the law of sines twice: sin(ABD)/AD = sin(20)/BD sin(180-ADB)/BC = sin(80)/BD Eliminate BD and note that 180 - ADB = 20 + ABD. Also, recall that AD=BC. All of this gives sin(20)*sin(ABD+20) = sin(ABD)*sin(80). Solve this for ABD. Then DBC = 80-ABD. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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