calculate the angles of a triangle?

Question

If we have a triangle $ABC$ and we only know three things:

-The angle $A$

-The length $AB$

-The length $AC$

Is it possible to calculate the other angles: $B$, $C$

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All what I can think of is:

$$ \frac{sin(B)}{AC} = \frac{sin(C)}{AB} $$

But since there are two unkowns, I think that this is not possible. Does anybody here has a way to do that? Or is the answer simply no?

Answer 1

Let $a = \angle BAC$ and $(b,c) = (|AB|,|AC|)$

Method 1

Let $D$ be on $AB$ such that $AB \perp CD$

Then $|AD| = c \cos(a)$

And $|CD| = c \sin(a)$

Thus $\angle ABC = \tan^{-1}(\frac{|CD|}{|AD|}) = \tan^{-1}(\frac{c\sin(a)}{b-c\cos(a)})$

Method 2

$|BC|^2 = b^2 + c^2 - 2 b c \cos(a)$

And $\frac{c}{\sin(\angle ABC)} = \frac{|BC|}{\sin(a)}$

Therefore $\angle ABC = \sin^{-1}\left(\frac{c\sin(a)}{\sqrt{b^2+c^2-2bc\cos(a)}}\right)$

Question

Why are the two formulae so different? Can you prove that they are equal without using this problem?