Question

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 1

Let the sides of the $△ABC$ be $a=n,b=n+1,c=n+2$, where n is natural number. Then C is the greatest and A the least angle.
As given C = 2A
$\therefore sinC=sin2A=2sinAcosA$
$\therefore kc=2ka\frac{b^2+c^2-a^2}{2bc}$
or $b^2=a(b^2+c^2-a^2).$
Substituting the values of a, b, c, we get
$(n+1)(n+2{)}^2=n[(n+1{)}^2+(n+2{)}^2-n^2]$
or $(n+1)(n+2{)}^2=n[n^2+6n+5]=n(n+1\left)\right(n+5)$
Since $n\ne -1$, we can cancel n+1
Thus $(n+2{)}^2=n(n+5)$
or $n^2+4n+4=n^2+5n$.
This gives n=4
Hence the sides are $4,5$ and $6$.