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Value of Angle DBC

Date: 1/23/96 at 21:30:58
From: aviel

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.


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!   
Associated Topics:
High School Trigonometry

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