Look at the pictures.
Method 1.
Use triangles from picture 1 and 2.
The triangleABC is isosceles triangle. Therefore [tex]|AB| = |BC| = 6[/tex].
[tex]|AC|=6\sqrt2=\dfrac{x\sqrt3}{2}\\\\\dfrac{x\sqrt3}{2}=6\sqrt2\ \ \ |\cdot2\\\\x\sqrt3=12\sqrt2\ \ \ |\cdot\sqrt3\\\\3x=12\sqrt6\ \ \ |:3\\\\x=4\sqrt6[/tex]
Answer: C. 4√6
Method 2
Use the trigonometric functions:
[tex]\sin45^o=\dfrac{6}{|AC|}\ \ \ /\sin45^o=\dfrac{\sqrt2}{2}/\\\\\dfrac{6}{|AC|}=\dfrac{\sqrt2}{2}\ \ \ |cross\ multiply\\\\|AC|\sqrt2=2\cdot6\ \ \ |\cdot\sqrt2\\\\2|AC|=12\sqrt2\ \ \ |:2\\\\|AC|=6\sqrt2[/tex]
[tex]\sin60^o=\dfrac{6\sqrt2}{x}\ \ \ /\sin60^o=\dfrac{\sqrt3}{2}/\\\\\dfrac{6\sqrt2}{x}=\dfrac{\sqrt3}{2}\ \ \ |cross\ multiply\\\\x\sqrt3=6\sqrt2\cdot2\ \ \ \ |\cdot\sqrt3\\\\3x=12\sqrt6\ \ \ \ |:3\\\\x=4\sqrt6[/tex]
