The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest…
- $3,4,5$
- $4,5,6$
- $5,6,7$
- $2,3,4$
Solution
Let $\mathrm{AC}=\mathrm{n}, \mathrm{AB}=\mathrm{n}+1, \mathrm{BC}=\mathrm{n}+2$
$\therefore \quad$ Largest angle is $\mathrm{A}$ and smallest angle is $\mathrm{B}$.
$\therefore \quad \mathrm{A}=2 \mathrm{~B}$
Since $\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ}$
$\therefore \quad 3 \mathrm{~B}+\mathrm{C}=180^{\circ}$
$\begin{aligned}
& \Rightarrow \mathrm{C}=180^{\circ}-3 \mathrm{~B} \\
& \Rightarrow \sin \mathrm{C}=\sin \left(180^{\circ}-3 \mathrm{~B}\right)=\sin 3 \mathrm{~B}
\end{aligned}$
By sine rule,
$\begin{aligned}
& \frac{\sin A}{n+2}=\frac{\sin B}{n}=\frac{\sin C}{n+1} \\
& \Rightarrow \frac{\sin 2 B}{n+2}=\frac{\sin B}{n}=\frac{\sin 3 B}{n+1} \\
& \Rightarrow \frac{2 \sin B \cos B}{n+2}=\frac{\sin B}{n}=\frac{3 \sin B-4 \sin ^3 B}{n+1} \\
& \Rightarrow \frac{2 \cos B}{n+2}=\frac{1}{n}=\frac{3-4 \sin ^2 B}{n+1} \\
\therefore & \cos B=\frac{n+2}{2 n}, 3-4 \sin ^2 B=\frac{n+1}{n} \\
\therefore \quad & 3-4\left(1-\cos ^2 B\right)=\frac{n+1}{n} \\
\therefore \quad & -4+4\left(\frac{n+2}{2 n}\right)^2=\frac{n+1}{n} \\
& \Rightarrow-1+\frac{n^2+4 n+4}{n^2}=\frac{n+1}{n} \\
& \Rightarrow-n^2+n^2+4 n+4=n^2+n \\
& \Rightarrow n^2-3 n-4=0 \\
& \Rightarrow(n+1)(n-4)=0 \\
& \Rightarrow n=-1 \text { or } n=4
\end{aligned}$
But $\mathrm{n}$ cannot be negative.
$\therefore \mathrm{n}=4$
$\therefore \quad$ The sides of the $\Delta$ are $4,5,6$.Asked in: MHT CET 2023 (13 May Shift 1)