Number of triangles $\Delta ABC$ with $\angle{ACB} = 30^o$ and $AC=9\sqrt{3}$ and $AB=9$? 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?
 A: Note that $$\sin (\angle ABC)=\frac{\sqrt3}2\quad\Rightarrow \angle ABC= 60^{\circ}\quad\text{or}\quad120^{\circ}$$Hence there is another triangle with angles $30^{\circ},30^{\circ},120^{\circ} $.
A: When you solved the Law of Sines equation, you forgot one solution.
Note that $$\sin (\angle ABC)=\frac{\sqrt3}{2}$$ implies that
$$\angle ABC= 60^{\circ}\quad\text{or}\quad120^{\circ}$$,
instead of just $60$.

I hope this helps.
A: Sketching the diagram systematically (and more reasonably; e.g., $AC$ should  be sketched almost twice as long as, instead of approximately the same length as, $AB$) helps the multiple cases become visible:


*

*In general, for $\theta\in(0^\circ,180^\circ),$
$$\sin\theta=k\implies\theta=\arcsin k \;\text{ or }\;
   180^\circ-\arcsin k,$$ while $$\cos\theta=k\implies\theta=\arccos
   k.$$



*Alternatively, using the Law of Cosines instead of the Law of Sines:
$$AB^2=AC^2+BC^2-2(AC)(BC)\cos\measuredangle{ACB}\\
   BC^2-27BC-162=0\\
   BC=9 \;\text{or}\; 18.$$
A: An alternative:
Consider,

$h$ is the height and it is equal to $a\sin\theta$.

*

*If $b\gt h$, there are two possible triangles.


 



*If $b=h$, there is one possible triangle.


 



*If $b\lt h$, there are no possible triangles.


 

