Answer:
(4[tex]\sqrt{3}[/tex])/3
Step-by-step explanation:
From the diagram,
Applying
sinθ = Opposite/Hypotenuse................. Equation 1
From the diagram,
Given: θ = 60°, Opposite = 2, Hypotenuse = X
Substitute these values into equation 1
sin60° = 2/X
make X the subject of the equation
X = 2/sin60°.................. Equation 2
But,
sin60° = [tex]\frac{\sqrt{3} }{2}[/tex]
Therefore,
X = 2/( [tex]\frac{\sqrt{3} }{2}[/tex])
X = (2×2)/[tex]\sqrt{3}[/tex]
X = 4/[tex]\sqrt{3}[/tex]
Rationalising the denominator,
X = (4/[tex]\sqrt{3}[/tex])×([tex]\sqrt{3}[/tex]/[tex]\sqrt{3}[/tex])
X = (4[tex]\sqrt{3}[/tex])/3