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### Value of Angle DBC

```
Date: 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:

Eliminate BD and note that 180 - ADB = 20 + ABD.  Also, recall that

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/
```
Associated Topics:
High School Trigonometry

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