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Equation of Circle in Complex Form

If A1,A2,...A...

Question

If $A_{1}, A_{2}, . . . A_{n}$ are vertices of regulr polygen of $n$ sides, inscribed in circle of radius $r$ , whose cnetre is origin $O$ , any $P$ is any point on the $a r c A_{n}, A_{1}$ such that $∠ P O A_{1} = θ$ . The value of sum of lenghts of the lines joining $P$ to the angluar points of the polygen is

A

2 r . sec \frac{π}{2 n} . sin (\frac{θ}{2} + \frac{π}{2 n})

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B

2 r . c o s e c \frac{π}{2 n} . cos (\frac{θ}{2} + \frac{π}{2 n})

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C

2 r . c o s e c \frac{π}{2 n} . cos (\frac{θ}{2} - \frac{π}{2 n})

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D

2 r . sec \frac{π}{2 n} . sin (\frac{θ}{2} - \frac{π}{2 n})

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Solution

The correct option is C $2 r . c o s e c \frac{π}{2 n} . cos (\frac{θ}{2} - \frac{π}{2 n})$
According to problem,

$∠_{1} O A_{2} = ∠ A_{2} O A_{3} = . . . ∠ A_{n} O A_{0} = \frac{2 π}{n}$

$∠ P O A_{1} = θ$ (Given)
$∠ P O A_{2} = θ + \frac{2 π}{n}$
$∠ P O A_{3} = θ + \frac{2 π}{n} . . .$
Hence if $r$ be the radius of the circle, we have
$P A_{1} = 2 r sin \frac{P O A_{1}}{2} = 2 r sin \frac{θ}{2}$
$P A_{2} = 2 r sin \frac{P O A_{2}}{2} = 2 r sin (\frac{θ}{2} + \frac{π}{n})$
Hence the required sum
$= 2 r [sin \frac{θ}{2} + sin (\frac{θ}{2} + \frac{π}{n}) + sin (\frac{θ}{2} + \frac{2 π}{n}) + . . . n terms]$
$= \frac{2 r [sin \frac{θ}{2} + (\frac{n - 1}{2}) \frac{π}{n} sin \frac{n π}{2 n}]}{sin \frac{π}{2 n}}$
$= 2 r c o s e c \frac{π}{2 n} sin [\frac{π}{2} + \frac{θ}{2} - \frac{π}{2 n}]$
$= 2 r c o s e c \frac{π}{2 n} cos (\frac{θ}{2} - \frac{π}{2 n})$

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