Question

The sides of a triangle are consecutive integers $n,n+1$ and $n+2$ and the largest angles is twice the smallest angle. Find $n$.

Answer 1

Let the smallest $\angle A=\theta$ thne the gratest $\angle C=2\theta$

In $△ABC$ by applying sine rule we get

$\frac{\sin \theta }{n}=\frac{\sin 2\theta }{n+2}$

$\frac{\sin \theta }{n}=\frac{2\sin \theta \cos \theta }{n+2}$

$\frac{1}{n}=\frac{2\cos \theta }{n+2}$

$\cos \theta =\frac{n+2}{2n}.............\left(1\right)$

In $△ABC$ by applying cosine rule we get

$\cos \theta =\frac{(n+1{)}^2+(n+2{)}^2-n^2}{2(n+1)(n+2)}........\left(2\right)$

Comparing (1) and (2) we get

$\frac{(n+1{)}^2+(n+2{)}^2-n^2}{2(n+1)(n+2)}=\frac{n+2}{2n}$

$(n+2{)}^2(n+1)=n(n+2{)}^2+n(n+1{)}^2-n^3$

$n(n+2{)}^2(n+2{)}^2=n(n+2{)}^2+n(n+1{)}^2-n^3$

$n^2+4n+4=n^3+2n^2+n-n^3$

$n^2-3n-4=0$

$(n+1)(n-4)=0$

$n=4(asn\ne -1)$

$\therefore$ Sides of triangle are $4,4+1,4+2$ n is $4,5,6$.