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Let A, B, C, D, E represent vertices of ...

Let A, B, C, D, E represent vertices of a regular pentagon ABCDE. Given the position vector of these vertices be `veca, veca + vecb, vecb, lamda veca and lamda vecb`, respectively.
The ratio `(AD)/(BC)` is equal to

A

`1-cos""(3pi)/(5) : cos""(3pi)/(5)`

B

`1+ 2cos""(2pi)/(5): cos ""(pi)/(5)`

C

`1 + 2 cos""(pi)/(5) : 2 cos""(pi)/(5)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
Given ABCDE is a regular pentagon

Let position vector of point A and C be `veca and veb`, respectively.
`AD` is parallel to BC and AB is parallel to EC.
Therefore,
AOCB is a parallelogram
and position vector of B is `veca + vecb`
The position vectors of E and D are `lamdavecb and lamda veca` respectively.
Also `OA = BC = AB = OC =1` (let)
Therefore, AOCB is rhombus.
`angle ABC = angle AOC = (3pi)/(5)`
and `angle OAB = angle BCO = pi - (3pi)/(5) = (2pi)/(5)`
Further `OA = AE = 1 and OC = CD = 1 `
Thus, `DeltaEAO and Delta OCD` are isosceles.
In `Delta OCD`, using sine rule we get `(OC)/(sin""(2pi)/(5))= (OD )/( sin""(pi)/(5))`
`rArr OD = (1)/( 2 cos ""(pi)/(5)) = OE`
`rArr AD = OA + OD = 1 + (1)/(2 cos ""(pi)/(2))`
`rArr (AD)/(BC) =1 + (1)/(2 cos""(pi)/(5)) = (1 +2 cos""(pi)/(5))/( 2 cos""(pi)/(5))`
And `(OE)/( OC) = (1)/( 2cos ""(pi)/(5))`
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