The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one determine the sides of the triangle

Answer

`4, 5, 6`

Explanation

Let `a=n, b=n+1, c=n+2, n in N`
Let the smallest angle `angleA=theta`, then the greatest angle `angleC=2theta`.

In `DeltaABC` by applying the sine law, we get
`(sin theta)/(n)=(sin 2 theta)/(n+2)`
or `(sin theta)/(n)=(2 sin theta cos theta)/(n+2)`
or `1/n=(2 cos theta)/(n+2)" "["as "sin theta ne 0]`
`implies cos theta=(n+2)/(2n)` (i)
In `DeltaABC`, by the cosine law, we get
`cos theta=((n+1)^(2)+(n+2)^(2)-n^(2))/(2(n+1)(n+2))` (ii)
Comparing the value of `cos theta` from Eqs. (i) and (ii), we get
`((n+1)^(2)+(n+2)^(2)-n^(2))/(2(n+1)(n+2))=(n+2)/(2n)`
or `(n+2)^(2) (n+1)=n(n+2)^(2)+n(n+1)^(2)-n^(3)`
or `n(n+2)^(2)+(n+2)^(2)=n(n+2)^(2)+n(n+1)^(2)-n^(3)`
or `n^(2)+4n+4=n^(3)+2n^(2)+n-n^(3)`
or `n^(2)-3n-4=0`
or `(n+1)(n-4)=0`
or `n-4" "["as "n ne -1]`
Therefore, the sides of the triangle are `4, 4+1, 4+2, i.e., 4, 5, 6`.