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In DeltaABC,(cosA)/(a)=(cosB)/(b)=(cosC)...

In `DeltaABC,(cosA)/(a)=(cosB)/(b)=(cosC)/(c)`. If `a=(1)/(sqrt(6))`, then the area of the triangle (in sq unit )is

A

`1//24`

B

`sqrt(3)//24`

C

`(1)/(8)`

D

`(1)/(sqrt(3))`

Text Solution

Verified by Experts

The correct Answer is:
B

Given , `(cosA)/(ksinA)=(cosB)/(ksinB)=(cosC)/(k sinC)`
`rArrcotA=cotB=cotC`
`rArrA=B=C=60^(@)`
`rArrDeltaABC` is an equilateral triangle
`thereforeDelta=(sqrt(3))/(4)a^(2)=(sqrt(3))/(4)xx(1)/(6)=(sqrt(3))/(24)`sq , unit
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