If-the-sides-of-a-triangle-are-consecutive-integers-and-the-maximum-angle-is-twice-the-minimum-determine-the-sides-of-the-triangle-

Question Number 76112 by Maclaurin Stickker last updated on 23/Dec/19

I f t h e s i d e s o f a t r i a n g l e a r e c o n s e c u t i v e

i n t e g e r s a n d t h e m a x i m u m a n g l e

i s t w i c e t h e m i n i m u m, d e t e r m i n e

t h e s i d e s o f t h e t r i a n g l e .

Answered by mr W last updated on 24/Dec/19

s a y t h e s i d e s a r e n - 1, n, n + 1

\cos θ_{m a x} = \frac{{(n - 1)}^{2} + n^{2} - {(n + 1)}^{2}}{2 (n - 1) n} = \frac{n - 4}{2 (n - 1)}

\cos θ_{m i n} = \frac{{(n + 1)}^{2} + n^{2} - {(n - 1)}^{2}}{2 (n + 1) n} = \frac{n + 4}{2 (n + 1)}

θ_{m a x} = 2 θ_{m i n}

\cos θ_{m a x} = 2 \cos^{2} θ_{m i n} - 1

\frac{n - 4}{2 (n - 1)} = 2 \times \frac{{(n + 4)}^{2}}{4 {(n + 1)}^{2}} - 1

\frac{n - 4}{n - 1} = \frac{4 n + 14 - n^{2}}{n^{2} + 2 n + 1}

2 n^{3} - 7 n^{2} - 17 n + 10 = 0

(n + 2) (n - 5) (2 n - 1) = 0

\Rightarrow n = 5

\Rightarrow s i d e s a r e 4, 5 a n d 6 .

Commented by Maclaurin Stickker last updated on 24/Dec/19

I d i d b y t h e s a m e w a y, b u t I {w a s n}^{'} t

s u r e . T h a n k y o u f o r t h e g r e a t

a n s w e r .