If the lengths of the sides of a triangle are in AP and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is

Answer

B

Explanation

Let a, b and c be the lengths of sides of a `DeltaABC` such that `a lt b lt c`.
Since, sides are in AP.
`therefore 2b = a + c " ...(i) "`
Let `angleA = theta`
Then, `angleC = 2 theta " [according to the equation]"`
So, `angleB = pi - 3 theta " (ii)"`
On applying sine rule in Eq. (i), we get
2 sin B = sin A + sin C
`implies 2 sin (pi - 3 theta) = sin theta + sin 2 theta " [from Eq. (ii)]"`
`implies 2 sin 3 theta = sin theta + sin 2 theta`
`implies 2 [3sin theta - 4 sin^(3) theta]= sin theta + 2sin theta cos theta`
`implies6 - 8sin^(2) theta = 1 + 2 cos theta" "[because sin theta " can not be zero]"`
`implies 6 - 8(1 - cos^(2)theta) = 1 + 2cos theta`
`implies 8 cos^(2) theta - 2 cos theta - 3 = 0`
`implies (2 cos theta + 1)(4 cos theta - 3) = 0`
`implies cos theta = (3)/(4)`
or `cos theta = - (1)/(2)` (rejected).
Clearly, the ratio of sides is a : b : c
`=sin theta : sin 3 theta : sin 2 theta`
`=sin theta : (3sin theta - 4 sin^(3) theta) : 2 sin theta cos theta`
`=1 : (4 cos^(2) theta - 1) : 2 cos theta`
`=1 : (5)/(4):(6)/(4) = 4 : 5: 6`