I came across the following question just now,
A triangle $\Delta ABC$ is drawn such that $\angle{ACB} = 30^o$ and side length $AC$ = $9*\sqrt{3}$
If side length $AB = 9$, how many possible triangles can $ABC$ exist as?
Here is a diagram for reference:
Here is what I did:
- I used the Law of Sines to find angle $\angle ABC$
$\to \frac{9}{\sin(30^o)} = \frac{9*\sqrt{3}}{\sin(\angle ABC)}$
$$\to \angle ABC = 60^o$$
So, therefore, $\Delta ABC$ can only exist as a $1$ triangle with angles: $30^o, 60^o$ and $90^o$.
But the answer says $2$ triangles are possible. So my question is: what is the second possible triangle?