In a triangle, the lengths of the two larger sides are 10 and 9, respectively. If the angles are in A.P., then the length of the third side can be `5-sqrt(6)` (b) `3sqrt(3)` (c) `5` (d) `5+sqrt(6)`

Answer

`5 + sqrt6 " or " 5 - sqrt6`

Explanation

Given that the angle of triangle are in A.P.,
Let `angleA = x - d, angle B = x, angle C = x + d`
Now, `angle A + angle B + angle C = 180^(@)`
`rArr x - d + x + x + d = 180^(@)`
`rArr 3x =180^(@)`
`rArr x = 60^(@)`
`:. Angle B = 60^(@)`
Using cosine formula `cos B = (a^(2) + c^(2) -B^(2))/(2AC)`, we get
`cos 60^(@) = (100 + c^(2) - 81)/(2 xx 10 xx c)`
`rArr (1)/(2) = (19 + c^(2))/(2 xx 10c)`
`rArr c^(2) - 10 c + 19 = 0`
`rArr c = (10 +- sqrt(100 - 76))/(2)`
`rArr c = 5 +- sqrt6`
Given that `a = 10 and b = 9` are the longer sides
Therefore, `c = 5 + sqrt6 and 5 - sqrt6` both are possible