triangle inscribed in a semi-circle

skeeter

MHF Helper
Jun 2008
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North Texas
You were given the tangent value for angle ABD ... from that, what did you get for the cosine and sine values for that angle?
 

Plato

MHF Helper
Aug 2006
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Yes, I understand that. It makes perfect sense for $m,~n,~p~\&~q$.
But it makes no sense for the rest of the question. $\angle BAD$ is not a right angle.
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
Yes, I understand that. It makes perfect sense for $m,~n,~p~\&~q$.
But it makes no sense for the rest of the question. $\angle BAD$ is not a right angle.
Agree, however, $\angle BDC$ is a right angle.

The structure of the problem was strange in that it implied knowing $\sin(\angle ABD)$ leads directly toward determining the length of BC.

I had to find $\cos(\angle DBC)$ (which it didn't ask for) to determine the length of BC. Once that was determined it was rather straightforward to finish.

Someone may have a better way ...
 
Jun 2013
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Lebanon
\(\displaystyle \frac{\text{AD}}{\sin \text{ABD}}=\text{BC}\)

Law of sines
 
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