Find the side lengths of a triangle if side lengths are consecutive integers,and one of whose angles is twice as large as another.


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Question Number 18968 by chux last updated on 02/Aug/17

$Find the side lengths of a triangle$ $if side lengths are consecutive$ $integers, and one of whose angles$ $is twice as large as another .$
Commented by chux last updated on 02/Aug/17

$please help$
	Answered by mrW1 last updated on 03/Aug/17

	$a = n - 1$ $b = n$ $c = n + 1$ $case 1 :$ $A = α$ $C = 2 α$ $B = 180 - 3 α$ $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ $\frac{\sin α}{n - 1} = \frac{\sin (3 α)}{n} = \frac{\sin (2 α)}{n + 1}$ $\frac{\sin (2 α)}{\sin α} = 2 \cos α = \frac{n - 1}{n + 1}$ $\frac{\sin (3 α)}{\sin α} = \frac{3 \sin α - 4 \sin^{3} α}{\sin α} = 3 - {4 \sin}^{2} α = {(2 \cos α)}^{2} - 1 = \frac{n}{n - 1}$ ${(\frac{n + 1}{n - 1})}^{2} - 1 = \frac{n}{n - 1}$ $(\frac{n + 1}{n - 1} + 1) (\frac{n + 1}{n - 1} - 1) = \frac{n}{n - 1}$ $\frac{4}{n - 1} = 1$ $\Rightarrow n = 5$ $\Rightarrow sides of triangle = 4, 5, 6$ $case 2 :$ $\frac{\sin α}{n - 1} = \frac{\sin (2 α)}{n} = \frac{\sin (3 α)}{n + 1}$ $\frac{\sin (2 α)}{\sin α} = 2 \cos α = \frac{n}{n - 1}$ $\frac{\sin (3 α)}{\sin α} = {(2 \cos α)}^{2} - 1 = \frac{n + 1}{n - 1}$ $(\frac{n}{n - 1} - 1) (\frac{n}{n - 1} + 1) = \frac{n + 1}{n + 1}$ $\frac{2 n - 1}{n - 1} = n + 1$ $\Rightarrow n = 2 (not suitable, {it}^{'} s a straight)$ $case 3 :$ $\frac{\sin (3 α)}{n - 1} = \frac{\sin α}{n} = \frac{\sin (2 α)}{n + 1}$ $\frac{\sin (2 α)}{\sin α} = 2 \cos α = \frac{n + 1}{n}$ $\frac{\sin (3 α)}{\sin α} = {(2 \cos α)}^{2} - 1 = \frac{n - 1}{n}$ $(\frac{n + 1}{n} - 1) (\frac{n + 1}{n} + 1) = \frac{n - 1}{n}$ $\Rightarrow n = - 2 < 0 (not suitable)$ $case 4 :$ $\frac{\sin (3 α)}{n - 1} = \frac{\sin (2 α)}{n} = \frac{\sin α}{n + 1}$ $\frac{\sin (2 α)}{\sin α} = 2 \cos α = \frac{n}{n + 1}$ $\frac{\sin (3 α)}{\sin α} = {(2 \cos α)}^{2} - 1 = \frac{n - 1}{n + 1}$ $(\frac{n}{n + 1} - 1) (\frac{n}{n + 1} + 1) = \frac{n - 1}{n + 1}$ $\Rightarrow n = - 2 < 0 (not suitable)$ $case 5 :$ $\frac{\sin (2 α)}{n - 1} = \frac{\sin α}{n} = \frac{\sin (3 α)}{n + 1}$ $\frac{\sin (2 α)}{\sin α} = 2 \cos α = \frac{n - 1}{n}$ $\frac{\sin (3 α)}{\sin α} = {(2 \cos α)}^{2} - 1 = \frac{n + 1}{n}$ $(\frac{n - 1}{n} + 1) (\frac{n - 1}{n} - 1) = \frac{n + 1}{n}$ $n = - 2 < 0 (not suitable)$ $case 6 :$ $\frac{\sin (2 α)}{n - 1} = \frac{\sin (3 α)}{n} = \frac{\sin α}{n + 1}$ $\frac{\sin (2 α)}{\sin α} = 2 \cos α = \frac{n - 1}{n + 1}$ $\frac{\sin (3 α)}{\sin α} = {(2 \cos α)}^{2} - 1 = \frac{n}{n + 1}$ $(\frac{n - 1}{n + 1} + 1) (\frac{n - 1}{n + 1} - 1) = \frac{n}{n + 1}$ $\frac{- 4}{n + 1} = 1$ $\Rightarrow n = - 5 < 0 (not suitable)$ $\Rightarrow the only solution is :$ $4, 5, 6$
	Commented by chux last updated on 03/Aug/17

	$but B = 3 α$
	Commented by ajfour last updated on 03/Aug/17

	$The same here . Thank you Sir .$
	Commented by chux last updated on 02/Aug/17

	$wow . . . . . this is amazing!$ $i m most grateful sir .$
	Commented by mrW1 last updated on 03/Aug/17

	$since we {don}^{'} t know which angle is$ $as double so large as which other angle,$ $we must try out all 6 possibilities :$ $A = α, B = 2 α, C = 180 - 3 α$ $A = α, B = 180 - 3 α, C = 2 α$ $A = 2 α, B = α, C = 180 - 3 α$ $A = 180 - 3 α, B = α, C = 2 α$ $A = 2 α, B = 180 - 3 α, C = α$ $A = 180 - 3 α, B = 2 α, C = α$ $note : \sin (180 - 3 α) = \sin (3 α)$
	Commented by chux last updated on 03/Aug/17

	$i^{'} ve understood it now . . . its clear$ $sir .$
	Commented by chux last updated on 04/Aug/17

	$thanks sir$
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