<s>[INST] Cosine rule problem Question:
In triangle ABC, BC = a, AC= b, AB = c, and BD is perpendicular to AC.
If $\angle ABC = 60^{\circ}$, prove that $c = \dfrac{1}{2}a \pm \sqrt{b^2-\dfrac{3}{4}a^2}$
My approach: $b^2 = a^2 + c^2 - 2ac\cos B$
$b^2 = a^2 + c^2 - 2ac \cos{60}^\circ$
$b^2 = a^2 + c^2 - ac$
$b^2 = (a-c)^2+ac$
After this I don't know how to isolate the $c$ term.
 [/INST] From $b^2 = a^2 + c^2 -ac$
$b^2 - a^2 = c^2 - ac$
$b^2 - \dfrac{3}{4}a^2 = c^2- ac + \dfrac{1}{4}a^2$
$b^2 - \dfrac{3}{4}a^2 = (c-\dfrac{1}{2}a)^2$
$c-\dfrac{1}{2}a = \pm \sqrt{b^2 - \dfrac{3}{4}a^2}$
$c = \dfrac{1}{2}a \pm \sqrt{b^2 - \dfrac{3}{4}a^2}$
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