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

Question

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?

Answer 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} $.

Answer 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 $60$.

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I hope this helps.

Answer 3

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:

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

Answer 4

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.