For this case we have the following trigonometric relationship:
[tex] cos(30)=\frac{CD}{2} [/tex]
From here, we clear the value of the CD segment:
[tex] CD = 2cos(30) [/tex]
[tex] CD = 2\frac{\sqrt{3}}{2} [/tex]
[tex] CD = \sqrt{3} [/tex]
Then, since the figure is a rectangle, then it follows that:
[tex] AB = CD = \sqrt{3} [/tex]
Answer:
the length of line segment AB is:
[tex] AB = \sqrt{3} [/tex]
option 2
The length of the side CD and AB is √3
[tex]Sin \theta=\dfrac{Perpendicular}{Hypotenuse}[/tex]
[tex]Cos \theta=\dfrac{Base}{Hypotenuse}[/tex]
[tex]Tan \theta=\dfrac{Perpendicular}{Base}[/tex]
where perpendicular is the side of the triangle which is opposite to the angle, and the hypotenuse is the longest side of the triangle which is opposite to the 90° angle.
Given to us
[tex]Sin \theta=\dfrac{Perpendicular}{Hypotenuse}[/tex]
[tex]Sin(\angle B)=\dfrac{CD}{BD}\\\\Sin(60^o)=\dfrac{CD}{2}\\\\CD = 2 \times Sin(60^o)\\\\ CD = 2 \times \dfrac{\sqrt3}{2}\\\\CD = \sqrt3[/tex]
Thus, the length of the side CD is √3.
As it is already given that the figure ABCD is a rectangle the length of side AB is equal to CD.
AB = CD = √3
Hence, the length of the side CD and AB is √3.
Learn more about Trigonometric functions:
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