Question

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

• A. $4,5,6$
• B. $5,6,7$
• C. $3,4,5$
• D. $7,8,9$

Answer 1

The correct option is A $4,5,6$

Let the sides of a triangle be $AC=n,AB=n+1,BC=n+2$ where $n\in N$
Smallest angle is $B$ and largest one is $A$
Given: $A=2B$
Also, $A+B+C={180}^{\circ }$
$3B+C={180}^{\circ }\Rightarrow C={180}^{\circ }-3B$
Now, using sine rule, $\frac{\sin A}{n+2}=\frac{\sin B}{n}=\frac{\sin C}{n+1}$
$\Rightarrow \frac{\sin 2B}{n+2}=\frac{\sin B}{n}=\frac{\sin ({180}^{\circ }-3B)}{n+1}$
$\Rightarrow \frac{\sin 2B}{n+2}=\frac{\sin B}{n}=\frac{\sin \left(3B\right)}{n+1}$
Now, $\frac{\sin 2B}{n+2}=\frac{\sin B}{n}\Rightarrow \frac{2\sin B.\cos B}{n+2}=\frac{\sin B}{n}\Rightarrow \cos B=\frac{n+2}{2n}\cdots \left(i\right)$
Now, $\frac{\sin B}{n}=\frac{\sin \left(3B\right)}{n+1}$
$\Rightarrow \frac{\sin B}{n}=\frac{\sin B(3-4{\sin }^{2}B)}{n+1}$
$\Rightarrow \frac{n+1}{n}=3-4(1-{\cos }^{2}B)\Rightarrow \frac{n+1}{n}=-1+4{\cos }^{2}B\cdots \left(ii\right)$
Now, from equation $\left(i\right)$ and $\left(ii\right)$, we get $\frac{n+1}{n}=-1+4{\left(\frac{n+2}{2n}\right)}^2$
$\Rightarrow \frac{n+1}{n}+1=\left(\frac{n^2+4n+4}{n^2}\right)$
$\Rightarrow \frac{2n+1}{n}=\frac{n^2+4n+4}{n^2}$
$\Rightarrow 2n^2+n=n^2+4n+4$
$\Rightarrow n^2-3n-4=0\Rightarrow (n-4)(n+1)=0$
So, we have $n=4$ or $n=-1$
Only possiblity is $n=4$
Hence the sides are $4,5,6$