Question

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

Let, the sides of the triangle be $a,b,c$, then $a+b+c=$

Answer 1

Let the sides of the triangle be $n,n+1,n+2$, if $A,B,C$ are the angles of the triangle where $A<B<C$

then $C=2A$ and $B=180-A-2A\Rightarrow B=80-3a$.

Thus, by the sine rule, we have

$\Rightarrow \frac{n}{\sin A}=\frac{n+1}{\sin 3A}=\frac{n+2}{\sin 2A}$

Considering $\frac{n}{\sin A}=\frac{n+2}{\sin 2A}$

$\Rightarrow \frac{sin2A}{\sin A}=\frac{n+2}{n}$

$\Rightarrow \frac{2sinAcosA}{\sin A}=\frac{n+2}{n}$

$\Rightarrow 2\cos A=\frac{n+2}{n}$

Squaring both sides. We get,

$\Rightarrow {\cos }^{2}A={\left(\frac{n+2}{2n}\right)}^2\dots \left(1\right)$

Now considering, $\frac{n}{\sin A}=\frac{n+1}{\sin 3A}$

$\Rightarrow \frac{sin3A}{\sin A}=\frac{n+1}{n}$

$\Rightarrow \frac{3\sin A-4{\sin }^{3}A}{\sin A}=\frac{n+1}{n}$

$3-4{\sin }^{2}A=\frac{n+1}{n}$

$\Rightarrow 3-4[1-{\left(\frac{n+2}{2n}\right)}^2]=\frac{n+1}{n}\dots \left(from\right(1\left)\right)$

$\Rightarrow 3-\frac{n+1}{n}=4[1-{\left(\frac{n+2}{2n}\right)}^2]$

$\Rightarrow \frac{2n-1}{n}=4\left[\frac{4n^2-{(n+2)}^2}{4n^2}\right]$

$\Rightarrow n(2n-1)=3n^2-4n-4\Rightarrow n^2-3n-4=0$

$\Rightarrow (n-4)(n+1)=0\Rightarrow n=4$ an $n$ is a natural number

Thus, sides of the triangle are $4,4+1,4+2\Rightarrow 4,5,6$

$\therefore a=4,b=5,c=6$

Thus, $a+b+c=15$