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Angle between Two Lines

Let A0 A1 A...

Question

Let $A_{0} A_{1} A_{2} A_{3} A_{4} A_{5}$ be a regular hexagon inscribed in a circle of unit radius. Find the product of the length of the line segments $A_{0} A_{1}, A_{1} A_{2}$ and $A_{0} A_{4}$ .

\sqrt{2}

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4

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\sqrt{3}

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6

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Solution

The correct option is C $\sqrt{3}$
As it is a regular hexagon , all the triangles are equilateral.
$O A_{0} = O A_{1} = A_{0} A_{1} = 1$
$O A_{1} = O A_{2} = A_{1} A_{2} = 1$
In $△ A_{4} O_{1} O, sin 6_{0} = \frac{A_{4} O_{1}}{A_{4} O} \Rightarrow A_{4} O_{1} = \frac{\sqrt{3}}{2}$
In $△ A_{0} O_{1} 0, A_{0} O_{1} = \frac{\sqrt{3}}{2}$

$A_{0} A_{4} = A_{4} O_{1} + A_{0} O_{1} = \sqrt{3}$

$= A_{0} A_{1} \times A_{1} A_{2} \times A_{0} A_{4} = 1 \times 1 \times \sqrt{3} = \sqrt{3}$

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Let A0 A1 A2 A3 A4 A5 be a regular hexagon inscribed in a circle of unit radius. Find the product of the length of the line segments A0 A1 , A1 A2 and A0 A4 .

The correct option is C √3As it is a regular hexagon , all the triangles are equilateral.OA0=OA1=A0A1=1OA1=OA2=A1A2=1 In △A4O1O,sin60=A4O1A4O⇒A4O1=√32In △A0O10,A0O1=√32A0A4=A4O1+A0O1=√3=A0A1×A1A2×A0A4=1×1×√3=√3

Let $A_{0} A_{1} A_{2} A_{3} A_{4} A_{5}$ be a regular hexagon inscribed in a circle of unit radius. Find the product of the length of the line segments $A_{0} A_{1}, A_{1} A_{2}$ and $A_{0} A_{4}$ .