Question

The sides of triangle are three consecutive natural numbers and its largest angle is twice the smaller one. The largest side of the triangle must be:

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is C $6$

Let $AB=n,AC=n+1,BC=n+2$

Further,
Let $A=2C$
Since $AB$ is the smallest and $BC$ is the largest.
By sine rule, we have
$\frac{\sin A}{n+2}=\frac{\sin B}{n+1}=\frac{\sin C}{n}$
$\because A=2C$ and $B={180}^0-(A+C)$
$\Rightarrow B={180}^0-3C$
$\Rightarrow \frac{2\sin C\cos C}{n+2}=\frac{3\sin C-4{\sin }^{3}C}{n+1}=\frac{\sin C}{n}$
On dividing by $\sin C$ we get
$\Rightarrow \frac{2\cos C}{n+2}=\frac{3-4{\sin }^{2}C}{n+1}=\frac{1}{n}$
$\therefore \cos C=\frac{n+2}{2n}$ and ${\sin }^{2}C=\frac{2n-1}{4n}$
Now, ${\sin }^{2}C+{\cos }^{2}C={\left(\frac{n+2}{2n}\right)}^2+{\left(\frac{2n-1}{4n}\right)}^2$
$\Rightarrow 1={\left(\frac{n+2}{2n}\right)}^2+{\left(\frac{2n-1}{4n}\right)}^2$
$\Rightarrow \frac{n^2+4n+4+n(2n-1)}{4n^2}=1$
$\Rightarrow n^2+4n+4+2n^2-n=4n^2$
$\Rightarrow n^2-3n-4=0$
$\therefore n=4,n=-1$
$\Rightarrow n=4$ and $n\ne -1\because n\in N$
Then sides are $4,5,6$
$\therefore$ Largest side$=6$