# triangle inscribed in a semi-circle

#### provasanteriores

somebody help me?

s=xy/z ??

#### skeeter

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

#### skeeter

MHF Helper
so, you're not going to share your results so that others may benefit?

1 person

#### Plato

MHF Helper
so, you're not going to share your results so that others may benefit?
AGREED! I still have no idea the point of the question.

#### skeeter

MHF Helper
AGREED! I still have no idea the point of the question.
all the boxed capital letters represent a base 10 digit

#### Plato

MHF Helper
all the boxed capital letters represent a base 10 digit
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
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 ...

#### Idea

$$\displaystyle \frac{\text{AD}}{\sin \text{ABD}}=\text{BC}$$

Law of sines

1 person