Number of triangles $\Delta ABC$ with $\angle{ACB} = 30^o$ and $AC=9\sqrt{3}$ and $AB=9$?

Solution 1:

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} $.

Solution 2:

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.

Solution 3:

An alternative:

Consider,

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

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

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

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

Solution 4:

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:

  $$AB^2=AC^2+BC^2-2(AC)(BC)\cos\measuredangle{ACB}\\ BC^2-27BC-162=0\\ BC=9 \;\text{or}\; 18.$$