The lengths of sides of a triangle are three consecutive natural numbers and its largest angle is twice the…
The lengths of sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Then the length of the sides of the triangle (in units) are
$3,4,5$
$4,5,6$
$5,6,7$
$2,3,4$
Solution
Let $a, a+1, a+2$ be the sides of the triangle and $\mathrm{A}, \mathrm{B}, \mathrm{C}$ be the angles opposite to them respectively.
According to the given condition,
$\mathrm{C}=2 \mathrm{~A}$
$\begin{aligned}
\therefore \quad \sin C & =\sin 2 A \\
\sin C & =2 \sin A \cos A
\end{aligned}$
Note that $\frac{\sin A}{a}=\frac{\sin C}{a+2}=k$
$\begin{aligned}
& \begin{array}{l}
\Rightarrow \sin A=k a \text { and } \sin C=k(a+2) \\
\text { Also, } \cos A=\frac{(a+1)^2+(a+2)^2-a^2}{2(a+1)(a+2)} \\
=\frac{a^2+2 a+1+a^2+4 a+4-a^2}{2\left(a^2+3 a+2\right)} \\
=\frac{a^2+6 a+5}{2\left(a^2+3 a+2\right)} \\
\therefore \quad(i) \Rightarrow k(a+2)=2 \times k a \times \frac{a^2+6 a+5}{2\left(a^2+3 a+2\right)} \\
\therefore \quad a+2=\frac{a\left(a^2+6 a+5\right)}{\left(a^2+3 a+2\right)} \\
\therefore \quad(a+2)\left(a^2+3 a+2\right)=a^3+6 a^2+5 a
\end{array}
\end{aligned}$
$\begin{array}{ll}
\therefore & a^3+5 a^2+8 a+4=a^3+6 a^2+5 a \\
\therefore & a^2-3 a-4=0 \\
\therefore & (a-4)(a+1)=0 \\
& \Rightarrow a=4 \text { or }-1
\end{array}$
But $\mathrm{a}=-1$ is not possible.
$\therefore \quad 4,5,6$ are the lengths of the sides of the triangle.