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?



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

enter image description here 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.

  • $\begingroup$ I get that you will get 2 answers from the graph, but what I don't understand is that there are two possible triangles, which means 2 different angle C values. How do I know there are two triangles? Or does the 0,1,2 solutions for the ambiguous law of sines not apply to this problem because it's not a SSA problem? $\endgroup$ – Hello Jun 16 '18 at 0:33
  • $\begingroup$ Be aware that while the length of $AC$ and $BC$ are the same in both triangles, $AB$ is different. Because the angle you've marked $\theta$ takes two different values, that will affect the other two angles and the length of $AB$. $\endgroup$ – Rhys Hughes Jun 16 '18 at 0:37
  • $\begingroup$ Wait but in the problem AC is 16 and BC is 6 $\endgroup$ – Hello Jun 16 '18 at 0:40
  • $\begingroup$ They are, yes that is correct. But the problem never said anything about the third side or the angles remaining the same. $\endgroup$ – Rhys Hughes Jun 16 '18 at 0:42
  • $\begingroup$ You can see them both drawn in steven's answer. $\endgroup$ – Rhys Hughes Jun 16 '18 at 11:50

enter image description here

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


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