In $\triangle ABC$, we have $AB = 14$, $BC = 16$, and $\angle A = 60^\circ$. Find the sum of all possible values of $AC$.

Question

In $\triangle ABC$, we have $AB = 14$, $BC = 16$, and $\angle A = 60^\circ$. Find the sum of all possible values of $AC$.

When I use the Law of Cosines, I get a quadratic like expected. However, when I use Vieta's Formulas to get the sum of the roots, I get the wrong answer. Please help me figure out what I did wrong.

Answer 1

What you did wrong was forgetting that there is a negative solution!

When you use the Law of Cosines, you get $$\begin{align}16^2 &= 14^2 + x^2 - 2(14)(x)cos(60^\circ) \\ 16^2 - 14^2 &= x^2 - 28x\cdot(\frac{1}{2}) \\ 16^2 - 14^2 &= x^2 - 14x \\ 60 &= x^2 - 14x \\ 0 &= x^2 - 14x - 60. \end{align}$$ Then we realize that one of the roots is negative leaving us with the answer $$\boxed{7 + \sqrt{109}.}$$

Answer 2

Using the law of cosines we have, $$\cos A=\frac{b^2+c^2-a^2}{2bc}$$ $$\cos\frac{\pi}{3}=\frac{b^2+196-256}{28b}$$ $$\frac12=\frac{b^2+196-256}{28b}$$ $$14b=b^2-60$$ $$b=7+\sqrt{109}$$ $$b\ne7-\sqrt{109}\mbox{ because the length cannot be in negative}$$