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

Let A,B,C,D,E represent vertices of a regular pentangon ABCDE. Given the position vector of these vertices be a,a+b,b,`lamda a` and `lamdab` respectively.
Q. AD divides EC in the ratio

A

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

B

`1+2"cos"(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 point A and C be a and b, respectively. AD is parallel to BC and AB is parallel to EC.
Therefore,
AOCB is a parallelogram annd position vector of B is a+b.
The position vectors of E annd D are `lamdab and lamda a` respectively.
Also, `OA=BC=AB=OC=1` (let)
therefore, `AOCB` is rhombus.
`angleABC=angleAOC=(3pi)/(5)`
and `angleOAB=angleBCO=pi-(3pi)/(5)=(2pi)/(5)`
further, `OA=AE=1 and OC=CD=1`
Thus, `DeltaEAO and DeltaOCD` are isosceles.
In `DeltaOCD`, using sine rule we get,
`(OC)/("sin"(2pi)/(5))=(OD)/("sin"(pi)/(5))`
`implies OD=(1)/(2"cos"(pi)/(5))=OE`
`implies AD=OA+OD=1+(1)/(2"cos"(pi)/(5))`
`(AD)/(BC)=1+(1)/(2"cos"(pi)/(5))=(1+2"cos"(pi)/(5))/(2"cos"(pi)/(5))`
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