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. 1, 2, 3
• B. 2,3,4
• C. 3,4,5
• D. 4,5,6

Answer 1

The correct option is D

4,5,6

Let the sides of $\Delta ABC$ be $a=n,b=n+1‘c=n+2,$ where $n$ is a natural number. Then $C$ is the greatest and $A$ the least angle. As given $C=2A$
$\therefore \sin C=\sin 2A=2\sin A\cos A\therefore kc=2ka\frac{b^2+c^2-a^2}{2bc}\text{or}b{c}^2=a(b^2+c^2-a^2)$
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 $n+1$.
$\text{Thus}(n+2{)}^2=n(n+5)\text{or}{n}^2+4n+4=n^2+5n$
This gives $n=4,$ hence the sides are 4, 5 and 6.