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The diagonals of a parallelogram are inc...

The diagonals of a parallelogram are inclined to each other at an angle of `45^@`, while its sides `a and b (a>0)` are inclined to each other at an angle of `30^@`, then the value of `a/b` is

A

`(3)/(2)`

B

`(3+sqrt(5))/(2)`

C

`(3+sqrt(5))/(4)`

D

`(sqrt(5)+1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
D


Let `AC=d_(1)` and `BD=d_(2)`
Area of parallelogram is
`(1)/(2)d_(1)d_(2)sin 45^(@)=2((1)/(2)ab sin 30^(@))`
`rArr d_(1)d_(2)=sqrt(2)ab`
where
`d_(1)^(2)=a^(2)+b^(2)-2ab cos 150^(@)=a^(2)+b^(2)+sqrt(3)ab`
and `d_(2)^(2)=a^(2)+b^(2)-2ab cos 30^(@)=a^(2)+b^(2)-sqrt(3) ab`
`rArr d_(1)^(2)d_(2)^(2)=(a^(2)+b^(2))^(2)-3a^(2)b^(2)`
`rArr 2a^(2)b^(2)=(a^(2)+b^(2))^(2)-3a^(2)b^(2)`
`rArr a^($)+b^(4)-3a^(2)b^(2)=0`
`rArr ((a)/(b))^(4)-3((a)/(b))^(2)+1=0`
`rArr ((a)/(b))^(2)=(3+sqrt(5))/(2)` (as a gt b)
`rArr ((a)/(b))^(2)=((sqrt(5)+1)^(2))/(4)`
`rArr (a)/(b)=(sqrt(5)+1)/(2)`
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