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

Options

A. `"cos"(2pi)/(5):1`
B. `"cos"(3pi)/(5):1`
C. `1:2"cos"(pi)/(5)`
D. `1:2`

Answer

C

Explanation

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))`
`(OE)/(OC)=(1)/(2"cos"(pi)/(5))`.