kiahtab13 kiahtab13
  • 22-06-2018
  • Mathematics
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‼️‼️Help anyone ‼️‼️‼️

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gmany
gmany gmany
  • 22-06-2018
Method 1
You can use The equilateral triangle (look at the picture)
[tex]\dfrac{a\sqrt3}{2}=1\ \ \ \cdot2\\\\a\sqrt3=2\ \ \ |\cdot\sqrt3\\\\3a=2\sqrt3\ \ \ |:3\\\\a=\dfrac{2\sqrt3}{3}\\\\\boxed{|AC|=\dfrac{2\sqrt3}{3}}[/tex]


Method 2
You can use the trigonometric function
[tex]\sin60^o=\dfrac{1}{|AC|}\\\\\sin60^o=\dfrac{\sqrt3}{2}\\\\\dfrac{1}{|AC|}=\dfrac{\sqrt3}{2}\ \ \ |\cross\ multiply\\\\|AC|\sqrt3=2\ \ \ |\cdot\sqrt3\\\\3|AC|=2\sqrt3\ \ \ \ |:3\\\\\boxed{|AC|=\dfrac{2\sqrt3}{3}}[/tex]

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