Ambiguous Case Triangle Problem

Question

1: The question: The area of triangle ABC is 24√3, side a = 6, and side b = 16. The value of angle C is

(A) 30°

(B) 30° or 150°

(C) 60°

(D) 60° or 120°

(E) None of the above

The book says the answer is D, but how do you know for certain that you can form 2 triangles? I tried using the a>h, a less than b method to see if there are two triangles but I can't because this isn't even a SSA problem. I get that when you use the area formula for a triangle, you get two answers for angle C: (1/2)*16*6sin(θ)=24√3 gets you θ=60° and 120°, but why can't answer (C) be correct? Couldn't there be only one triangle and thus you can only have one value for angle C?

1

Answer 1

The area of a triangle is given by the formula: $$\text{Area}=\frac12ab\sin C=\frac12ac\sin B=\frac12bc\sin A$$

In your case $$24\sqrt{3}=\frac 12(6)(16)\sin C$$ Rearranging we get: $$\sin C=\frac{\sqrt3}{2}$$ Using your calculator to evaluate this we get the principal root $C=60$.

However, looking at the sine graph, you can see that it will not only cross at $60^0$ but also at $120^0$.

Essentially, if the principal root is at $k$, where $k<180$, $180-k$ is also a root, as are $360+k$ and $540-k$ etc.

If $180<k<360$, the exact opposite is true, the roots are at $k, 360-k, 540+k, 720-k$ etc.

You can clearly see this by looking at a $\sin$ graph and seeing where the height from the $x$-axis is the same.

In the case of this question, the relevant fact is that $C=k$ and $C=180-k$ are both roots provided that $k<180$, which it is as $k=60$, so $180-60=120$ is also a root.

Answer 2

Because $B_1C=6$ and $h=3\sqrt 3$, $m\angle B_1CA = 120^\circ$ and $m\angle B_2CA = 60^\circ$