The sides of a triangle are in A.P. and its area is `3/5t h` of the an equilateral triangle of the same perimeter, prove that its sides are in the ratio 3:5:7.

Answer

`120^(@)`

Explanation

Let the sides be `x - d, x, x + d`. Then
`s = (3x)/(2), (s -a) = (x)/(2) + d`,
`(s-b) = (x)/(2), (s-c) = (x)/(2) -d`.
Area of triangle `= sqrt((3x)/(2).((x)/(2) + d).(x)/(2).((x)/(2) -d))`
`= (x)/(2) sqrt(3((x^(2))/(4) -d^(2))) = (x)/(4) sqrt(3(x^(2) -4d^(2)))`
The area of equilateral triangle whose perimeter is `3x " is " (sqrt3x^(2))/(4)`
Given, `(3)/(5) xx (sqrt3x^(2))/(4) = (x)/(4) sqrt(3(x^(2) -4d^(2)))`
or `(9)/(25) xx (3x^(4))/(16) = (x^(2))/(16) xx 3 (x^(2) -4d^(2))`
or `x^(2) - (9x^(2))/(25) = 4d^(2)`
or `16 x^(2) = 100d^(2)`
or `x = (5)/(2) d`
Therefore, the sides of triangle measure `((5d)/(2) -d), (5d)/(2), ((5d)/(2) + d)`
or `(3d)/(2), (5d)/(2), (7d)/(2)`
Hence, the ratio of sides is `3 : 5 : 7`
For the greatest angle,
`cos theta = (3^(2) + 5^(2) -7^(2))/(2 xx 3 xx 5) = (9 + 25 - 49)/(30) = (-1)/(2)`
`:. theta = 120^(2)`