JEE MainMathematicsToughProperties of triangle

The lengths of sides of a triangle are three consecutive natural
numbers and its largest angle is twice the smaller one. Perimeter of the
triangle is

Show answer and solution

Let the lengths of the sides be n, n + 1, n + 2 where n

\in

N.
From the question, the largest angle opposite to the side n + 2 is
2

\theta

while the smallest angle opposite to the side n is

\theta

.
Now,

cos\theta = \frac{(n + 2)^{2} + (n + 1)^{2} - n^{2}}{2(n + 1)(n + 2)} = \frac{n + 5}{2(n + 2)}

And

cos2\theta = \frac{n^{2} + (n + 1)^{2} - (n + 2)^{2}}{2n(n + 1)}

= + 1)}

But

cos2\theta = 2\cos^{2}\theta - 1;\ so\ \frac{n - 3}{2n} = 2\left\{ \frac{n + 5}{2(n + 2)} \right\}^{2} - 1

\frac{n - 3}{2n} = \frac{(n + 5)^{2}}{2(n + 2)^{2}} - 1

(n - 3)(n + 2)^{2} = n\left\{ (n + 5)^{2} - 2(n + 2)^{2} \right\}

(n - 3)\left( n^{2} + 4n + 4 \right) = n\left( - n^{2} + 2n + 17 \right)

or n^3 + n^2 - 8n - 12 = - n^2 + 2n^2 + 17n
or 2n^3 - n^2 - 25n-12 = 0
or (n - 4) (2n^2 + 7n + 3) = 0
or n = 4 or 2n^2 + 7n +3 =

Source: Grass-G / M-23 Mathematics Paper 31 Dec FST.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.