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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 andlamda vecb, respectively. AD divides EC in the ratio |
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Answer» `cos""(2pi)/(5) : 1` 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 `angleOAB = 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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