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Question Number 38094 by ajfour last updated on 21/Jun/18

Commented by ajfour last updated on 21/Jun/18

$A r o d h a n g s f r o m c e i l i n g b y t w o$ $t w o s t r i n g s o f l e n g t h a, b f r o m$ $p o i n t s o n c e i l i n g s e p a r a t e d b y$ $d i s t a n c e d . T h e r o d h a s l e n g t h l .$ $F i n d θ .$
Commented by MrW3 last updated on 23/Jun/18

$I m a g e t h a t w e c o u l d e x t e n d t h e s t r i n g s$ $i n t o t h e c e i l i n g, t h e n t h e y w o u l d m e e t$ $a t a p o i n t, {l e t}^{'} s s a y p o i n t O .$ $P o i n t O m u s t l i e o n t h e s a m e v e r t i c a l$ $l i n e a s t h e C O M o f t h e r o d, {l e t}^{'} s s a y$ $p o i n t M .$ $S o i n a f i r s t s t e p I w o u l d t r y t o s o l v e$ $t h e e a s i e r c a s e w h e r e t h e r o d h a n g s o n$ $t h e p o i n t O t h r o u g h t w o s t r i n g s w i t h$ $l e n g t h e s l_{1} a n d l_{2} . W e w i l l a l s o h a v e$ $t h i s c a s e i f d = 0 .$
Commented by MrW3 last updated on 23/Jun/18

Commented by MrW3 last updated on 23/Jun/18

$l e t l_{1} = O A, l_{2} = O B$ $w e k n o w O M i s a m e d i a n o f t h e t r i a n g l e$ $O A B a n d$ $O M = \frac{1}{2} \sqrt{2 (l_{1}^{2} + l_{2}^{2}) - l^{2}}$ $c o s ∠ O M B = \frac{- l_{2}^{2} + {O M}^{2} + {M B}^{2}}{2 \times O M \times M B}$ $s i n c e θ = 90 ° - ∠ O M B$ $\Rightarrow \sin θ = \frac{- l_{2}^{2} + {O M}^{2} + {M B}^{2}}{2 \times O M \times M B}$ $\Rightarrow \sin θ = \frac{- l_{2}^{2} + \frac{2 (l_{1}^{2} + l_{2}^{2}) - l^{2}}{4} + {(\frac{l}{2})}^{2}}{2 \times \frac{\sqrt{2 (l_{1}^{2} + l_{2}^{2}) - l^{2}}}{2} \times \frac{l}{2}}$ $\Rightarrow \sin θ = \frac{- 4 l_{2}^{2} + 2 (l_{1}^{2} + l_{2}^{2}) - l^{2} + l^{2}}{2 l \sqrt{2 (l_{1}^{2} + l_{2}^{2}) - l^{2}}}$ $\Rightarrow \sin θ = \frac{l_{1}^{2} - l_{2}^{2}}{l \sqrt{2 (l_{1}^{2} + l_{2}^{2}) - l^{2}}}$ $\Rightarrow θ = \sin^{- 1} \frac{l_{1}^{2} - l_{2}^{2}}{l \sqrt{2 (l_{1}^{2} + l_{2}^{2}) - l^{2}}}$ $\cos α = \frac{{O A}^{2} + {O M}^{2} - {A M}^{2}}{2 \times O A \times O M}$ $\cos α = \frac{l_{1}^{2} + \frac{2 (l_{1}^{2} + l_{2}^{2}) - l^{2}}{4} - {(\frac{l}{2})}^{2}}{2 \times l_{1} \times \frac{\sqrt{2 (l_{1}^{2} + l_{2}^{2}) - l^{2}}}{2}}$ $\cos α = \frac{3 l_{1}^{2} + l_{2}^{2} - l^{2}}{2 l_{1} \sqrt{2 (l_{1}^{2} + l_{2}^{2}) - l^{2}}}$ $\Rightarrow α = \cos^{- 1} \frac{3 l_{1}^{2} + l_{2}^{2} - l^{2}}{2 l_{1} \sqrt{2 (l_{1}^{2} + l_{2}^{2}) - l^{2}}}$ $\Rightarrow β = \cos^{- 1} \frac{3 l_{2}^{2} + l_{1}^{2} - l^{2}}{2 l_{2} \sqrt{2 (l_{1}^{2} + l_{2}^{2}) - l^{2}}}$
Commented by ajfour last updated on 23/Jun/18

$B r i l l i a n t t h o u g h t s i r,$ $b u t l_{1}, l_{2} a r e t h e m s e l v e s$ $d e p e n d e n t o n α, β; S i r . .$
Commented by MrW3 last updated on 23/Jun/18

$y e s .$ $t h e p r o b l e m i s s t i l l n o t e a s y t o s o l v e .$ $o n e c a n n o t g e t a c l o s e d f o r m u l a f o r$ $c a s e d \neq 0 t o c a l c u l a t e θ d i r e c t l y .$
Commented by ajfour last updated on 23/Jun/18

$T h a n k y o u a n y w a y S i r .$
Commented by MrW3 last updated on 24/Jun/18

$I {d o n}^{'} t t h i n k {i t}^{'} s p o s s i b l e t o f i n d a$ $c l o s e d f o r m u l a t o c a l c u l a t e t h e a n g l e$ $θ d i r e c t l y . I f w e k n o w t h e a n g l e ∠ B A C,$ ${l e t}^{'} s s a y φ, t h e n w e c a n a l s o f i n d θ .$ $I t h i n k w e c a n g e t a s i n g l e e q u a t i o n,$ $t h o u g h a c o m p l i c a t e d o n e, f o r φ,$ $s o m e t h i n g l i k e F (φ, a, b, d, l) = 0, n o t$ $i n f o r m o f φ = F (a, b, d, l) .$
Commented by ajfour last updated on 24/Jun/18

Commented by ajfour last updated on 24/Jun/18
${ d_1 +acos α=(l/2)cos θ }×tan 𝛂 {d_2 +bcos β =(l/2)cos θ }×tan β Since d_1 tan α = d_2 tan β =h asin α−bsin β =(l/2)cos θ(tan α−tan β) Now asin α−bsin β = lsin θ _____________________ ⇒ tan 𝛂−tan 𝛃 = 2tan 𝛉 _____________________ Considering the △ with sides a, b, and ρ we have (ρ/(sin (α+β)))=(a/(sin (β+δ)))=(b/(sin (α−δ))) α=sin^(−1) [((bsin (α+β))/ρ)]+δ β=sin^(−1) [((asin (α+β))/ρ)]−δ where (as can easily be seen) 𝛒^2 =(lcos 𝛉−d)^2 +l^2 sin^2 𝛉 tan 𝛅 = ((lsin 𝛉)/(lcos 𝛉−d)) cos (𝛂+𝛃)=((a^2 +b^2 −𝛒^2 )/(2ab)) sin (𝛂+𝛃)=(√(1−[((a^2 +b^2 −𝛒^2 )/(2ab))]^2 )) ρ^2 −b^2 sin^2 (α+β)=ρ^2 −b^2 +(((a^2 +b^2 −ρ^2 )^2 )/(4a^2 )) =((4a^2 (ρ^2 −b^2 )+[a^2 −(ρ^2 −b^2 )]^2 )/(4a^2 )) =[((a^2 +ρ^2 −b^2 )/(2a))]^2 ⇒ 2absin (α+β)=(√(4a^2 b^2 −(a^2 +b^2 −ρ^2 )^2 )) ⇒bsin (α+β)=((√([(a+b)^2 −ρ^2 ][ρ^2 −(a−b)^2 ]))/(2a)) ________________________ 2tan 𝛉 = tan α−tan β =tan {sin^(−1) [((bsin (α+β))/ρ)]+δ} −tan {sin^(−1) [((asin (α+β))/ρ)]−δ} =tan {tan^(−1) (((bsin (α+β))/(√(ρ^2 −b^2 sin^2 (α+β)))))+𝛅} −tan {tan^(−1) (((asin (α+β))/(√(ρ^2 −a^2 sin^2 (α+β))))−𝛅} __________________________ 2tan θ =tan [tan^(−1) ((√([(a+b)^2 −ρ^2 ][ρ^2 −(a−b)^2 ]))/(a^2 +ρ^2 −b^2 ))+δ] −tan [tan^(−1) ((√([(a+b)^2 −ρ^2 ][ρ^2 −(a−b)^2 ]))/(b^2 +ρ^2 −a^2 ))−δ] ___________________________ .$
${d_{1} + a \cos α = \frac{l}{2} \cos θ} \times \tan α$ ${d_{2} + b \cos β = \frac{l}{2} \cos θ} \times \tan β$ $S i n c e d_{1} \tan α = d_{2} \tan β = h$ $a \sin α - b \sin β = \frac{l}{2} \cos θ (\tan α - \tan β)$ $N o w a \sin α - b \sin β = l \sin θ$ $_____________________$ $\Rightarrow \tan α - \tan β = 2 \tan θ$ $_____________________$ $C o n s i d e r i n g t h e △ w i t h s i d e s$ $a, b, a n d ρ w e h a v e$ $\frac{ρ}{\sin (α + β)} = \frac{a}{\sin (β + δ)} = \frac{b}{\sin (α - δ)}$ $α = \sin^{- 1} [\frac{b \sin (α + β)}{ρ}] + δ$ $β = \sin^{- 1} [\frac{a \sin (α + β)}{ρ}] - δ$ $w h e r e (a s c a n e a s i l y b e s e e n)$ $ρ^{2} = {(l \cos θ - d)}^{2} + l^{2} \sin^{2} θ$ $\tan δ = \frac{l \sin θ}{l \cos θ - d}$ $\cos (α + β) = \frac{a^{2} + b^{2} - ρ^{2}}{2 a b}$ $\sin (α + β) = \sqrt{1 - {[\frac{a^{2} + b^{2} - ρ^{2}}{2 a b}]}^{2}}$ $ρ^{2} - b^{2} \sin^{2} (α + β) = ρ^{2} - b^{2} + \frac{{(a^{2} + b^{2} - ρ^{2})}^{2}}{4 a^{2}}$ $= \frac{4 a^{2} (ρ^{2} - b^{2}) + {[a^{2} - (ρ^{2} - b^{2})]}^{2}}{4 a^{2}}$ $= {[\frac{a^{2} + ρ^{2} - b^{2}}{2 a}]}^{2}$ $\Rightarrow 2 a b \sin (α + β) = \sqrt{4 a^{2} b^{2} - {(a^{2} + b^{2} - ρ^{2})}^{2}}$ $\Rightarrow b \sin (α + β) = \frac{\sqrt{[{(a + b)}^{2} - ρ^{2}] [ρ^{2} - {(a - b)}^{2}]}}{2 a}$ $________________________$ $2 \tan θ = \tan α - \tan β$ $= \tan {\sin^{- 1} [\frac{b \sin (α + β)}{ρ}] + δ}$ $- \tan {\sin^{- 1} [\frac{a \sin (α + β)}{ρ}] - δ}$ $= \tan {\tan^{- 1} (\frac{b \sin (α + β)}{\sqrt{ρ^{2} - b^{2} \sin^{2} (α + β)}}) + δ}$ $- \tan {\tan^{- 1} (\frac{a \sin (α + β)}{\sqrt{ρ^{2} - a^{2} \sin^{2} (α + β)}} - δ}$ $__________________________$ $2 \tan θ$ $= \tan [\tan^{- 1} \frac{\sqrt{[{(a + b)}^{2} - ρ^{2}] [ρ^{2} - {(a - b)}^{2}]}}{a^{2} + ρ^{2} - b^{2}} + δ]$ $- \tan [\tan^{- 1} \frac{\sqrt{[{(a + b)}^{2} - ρ^{2}] [ρ^{2} - {(a - b)}^{2}]}}{b^{2} + ρ^{2} - a^{2}} - δ]$ $___________________________.$
Commented by MrW3 last updated on 24/Jun/18

$t h i s e q n . i s c o r r e c t s i r . I c h e c k e d$ $d i f f e r e n t v a l u e s f o r a, b, d a n d l . a n d$ $t h e s o l u t i o n i s r i g h t .$ $w o n d e r f u l w o r k i n g!$
Commented by ajfour last updated on 25/Jun/18

$T h a n k s y o u S i r!$
	Answered by ajfour last updated on 24/Jun/18
	$2tan θ = tan {tan^(−1) [((√([(a+b)^2 −ρ^2 ][ρ^2 −(a−b)^2 ]))/(a^2 +ρ^2 −b^2 ))]+δ} −tan { tan^(−1) [((√([(a+b)^2 −ρ^2 ][ρ^2 −(a−b)^2 ]))/(b^2 +ρ^2 −a^2 ))]−δ} where 𝛒^2 = l^2 sin^2 𝛉+(lcos 𝛉−d)^2 and 𝛅 = tan^(−1) (((lsin 𝛉)/(lcos 𝛉−d))) .$
	$2 \tan θ$ $= t a n {\tan^{- 1} [\frac{\sqrt{[{(a + b)}^{2} - ρ^{2}] [ρ^{2} - {(a - b)}^{2}]}}{a^{2} + ρ^{2} - b^{2}}] + δ}$ $- t a n {\tan^{- 1} [\frac{\sqrt{[{(a + b)}^{2} - ρ^{2}] [ρ^{2} - {(a - b)}^{2}]}}{b^{2} + ρ^{2} - a^{2}}] - δ}$ $w h e r e$ $ρ^{2} = l^{2} \sin^{2} θ + {(l \cos θ - d)}^{2}$ $a n d δ = \tan^{- 1} (\frac{l \sin θ}{l \cos θ - d}) .$
	Commented by ajfour last updated on 24/Jun/18

	$T o p r o c e e d b e y o n d t h i s, {i s n}^{'} t$ $p o s s i b l e f o r m e . T h i s i s a l l i$ $c o u l d a r r i v e a t, b u t ∠ B A C o r / a n d$ $∠ A B D i s / a r e e l i m i n a t e d, e v e n t u a l l y .$
	Commented by MrW3 last updated on 24/Jun/18

	${T h a t}^{'} s g r e a t s i r! Y o u h a v e a t t h e e n d$ $o n e s i n g l e e q u a t i o n w i t h o n l y o n e$ $u n k n o w n . T h e e q n . i s e v e n m o r e s i m p l e$ $t h a n I^{'} v e e x p e c t e d .$
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