Another challenging question, no sine and no cosine law

  • Thread starter davedave
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Main Question or Discussion Point

This is the 3rd problem from the book, The Unsolvable and the Solvable.

This is the last one I post from this book. It is pretty challenging.

Once again, it is NOT a homework problem.

Consider an isoceles triangle ABC and an equilateral triangle BCD which share the side BC as shown below. IGNORE the dotted lines in the diagram.


.........................A


..................D
...................................E

..............B_______________C



..........................D


D is a point on side AB and E is a point on side AC.

angle DAE=20 degrees
angle DEA=20 degrees
angle EDC=10 degrees
angle ECD=10 degrees
angle DBC=80 degrees
angle DCB=70 degrees
angle BDC=30 degrees

WITHOUT using the sine and the cosine law, determine angle EDC
 

Answers and Replies

  • #2
CRGreathouse
Science Advisor
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equilateral triangle BCD
[...]
angle BDC=30 degrees
I'd say no solution.
 
  • #3
There are two points labeled D! This is not a well defined problem.
 

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