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?

• With the sides and angle given, the only possibilities are $\space\angle ABC=60^\circ,\space BC=18\space$ or $\space\angle ABC=120^\circ,\space BC=9. \space$ Dec 4 '21 at 2:40

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 $$30^{\circ},30^{\circ},120^{\circ}$$.

• Oh right. I forgot that sine is positive in quadrant 2 and the angle is less than $180^o$, Thanks a lot! Dec 2 '21 at 5:51
• You're welcome. I'm glad I could help. Dec 2 '21 at 5:58

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.

• It really does, thanks for helping :) Dec 2 '21 at 5:56

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.
1. If $$b=h$$, there is one possible triangle.
1. If $$b\lt h$$, there are no possible triangles.
• When $b > h$, there is always a second possibility for an angle? Dec 2 '21 at 6:06
• @ShootingStars ... yes, and here we are considering the cases where $\theta\lt90^\circ.$
– user997661
Dec 2 '21 at 7:21

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:

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

• This is a really good way of looking at it. I almost get your idea with LOC(Law of Cosines), however, where would you start? Would you use it to find the length of the third unknown side first as its given by the following equation right: $a^2 = b^2+c^2-2bc*\cos(\theta)$? Dec 2 '21 at 6:01
• Oh, my bad, I must have thought wrong Dec 2 '21 at 6:08
• You can apply the Law of Cosines here. By applying the equation $c^2 = a^2 + b^2 - 2ab\cos\gamma$ with $b = 9\sqrt{3}$, $b = 9$, and $\gamma = 30^\circ$, you obtain a quadratic equation in $a$. Since it has a positive discriminant, the triangle has two possible solutions. Dec 3 '21 at 11:00
• @N.F.Taussig Ah, your helpful observation makes clear that whenever ambiguity should arise from the Sine Law, there is a corresponding ambiguity (in the same triangle) arising from the Cosine Law. (While the Sine Law will not weed any any illegitimate angle—one that causes the triangle's angles to exceed $180^\circ$—the Cosine Law's negative discriminant does this job.) I'd never thought to use the Cosine Law this way before; thanks for pointing out! Dec 3 '21 at 15:54