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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 rod hangs from ceiling by two two strings of length a, b from points on ceiling separated by distance d. The rod has length l. Find 𝛉.

$${A}\:{rod}\:{hangs}\:{from}\:{ceiling}\:{by}\:{two} \\ $$$${two}\:{strings}\:{of}\:{length}\:\boldsymbol{{a}},\:\boldsymbol{{b}}\:{from} \\ $$$${points}\:{on}\:{ceiling}\:{separated}\:{by} \\ $$$${distance}\:\boldsymbol{{d}}.\:{The}\:{rod}\:{has}\:{length}\:\boldsymbol{{l}}. \\ $$$$\boldsymbol{{F}}{ind}\:\boldsymbol{\theta}. \\ $$

Commented by MrW3 last updated on 23/Jun/18

Image that we could extend the strings into the ceiling, then they would meet at a point, let′s say point O. Point O must lie on the same vertical line as the COM of the rod, let′s say point M. So in a first step I would try to solve the easier case where the rod hangs on the point O through two strings with lengthes l_1 and l_2 . We will also have this case if d=0.

$${Image}\:{that}\:{we}\:{could}\:{extend}\:{the}\:{strings} \\ $$$${into}\:{the}\:{ceiling},\:{then}\:{they}\:{would}\:{meet} \\ $$$${at}\:{a}\:{point},\:{let}'{s}\:{say}\:{point}\:{O}. \\ $$$${Point}\:{O}\:{must}\:{lie}\:{on}\:{the}\:{same}\:{vertical} \\ $$$${line}\:{as}\:{the}\:{COM}\:{of}\:{the}\:{rod},\:{let}'{s}\:{say} \\ $$$${point}\:{M}. \\ $$$${So}\:{in}\:{a}\:{first}\:{step}\:{I}\:{would}\:{try}\:{to}\:{solve} \\ $$$${the}\:{easier}\:{case}\:{where}\:{the}\:{rod}\:{hangs}\:{on} \\ $$$${the}\:{point}\:{O}\:{through}\:{two}\:{strings}\:{with} \\ $$$${lengthes}\:{l}_{\mathrm{1}} \:{and}\:{l}_{\mathrm{2}} .\:\:{We}\:{will}\:{also}\:{have} \\ $$$${this}\:{case}\:{if}\:{d}=\mathrm{0}. \\ $$

Commented by MrW3 last updated on 23/Jun/18

Commented by MrW3 last updated on 23/Jun/18

let l_1 =OA, l_2 =OB we know OM is a median of the triangle OAB and OM=(1/2)(√(2(l_1 ^2 +l_2 ^2 )−l^2 )) cos∠OMB=((−l_2 ^2 +OM^2 +MB^2 )/(2×OM×MB)) since θ=90°−∠OMB ⇒sin θ=((−l_2 ^2 +OM^2 +MB^2 )/(2×OM×MB)) ⇒sin θ=((−l_2 ^2 +((2(l_1 ^2 +l_2 ^2 )−l^2 )/4)+((l/2))^2 )/(2×((√(2(l_1 ^2 +l_2 ^2 )−l^2 ))/2)×(l/2))) ⇒sin θ=((−4l_2 ^2 +2(l_1 ^2 +l_2 ^2 )−l^2 +l^2 )/(2l(√(2(l_1 ^2 +l_2 ^2 )−l^2 )))) ⇒sin θ=((l_1 ^2 −l_2 ^2 )/(l(√(2(l_1 ^2 +l_2 ^2 )−l^2 )))) ⇒θ=sin^(−1) ((l_1 ^2 −l_2 ^2 )/(l(√(2(l_1 ^2 +l_2 ^2 )−l^2 )))) cos α=((OA^2 +OM^2 −AM^2 )/(2×OA×OM)) cos α=((l_1 ^2 +((2(l_1 ^2 +l_2 ^2 )−l^2 )/4)−((l/2))^2 )/(2×l_1 ×((√(2(l_1 ^2 +l_2 ^2 )−l^2 ))/2))) cos α=((3l_1 ^2 +l_2 ^2 −l^2 )/(2l_1 (√(2(l_1 ^2 +l_2 ^2 )−l^2 )))) ⇒α=cos^(−1) ((3l_1 ^2 +l_2 ^2 −l^2 )/(2l_1 (√(2(l_1 ^2 +l_2 ^2 )−l^2 )))) ⇒β=cos^(−1) ((3l_2 ^2 +l_1 ^2 −l^2 )/(2l_2 (√(2(l_1 ^2 +l_2 ^2 )−l^2 ))))

$${let}\:{l}_{\mathrm{1}} ={OA},\:{l}_{\mathrm{2}} ={OB} \\ $$$${we}\:{know}\:{OM}\:{is}\:{a}\:{median}\:{of}\:{the}\:{triangle} \\ $$$${OAB}\:{and} \\ $$$${OM}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}\left({l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} \right)−{l}^{\mathrm{2}} } \\ $$$${cos}\angle{OMB}=\frac{−{l}_{\mathrm{2}} ^{\mathrm{2}} +{OM}^{\mathrm{2}} +{MB}^{\mathrm{2}} }{\mathrm{2}×{OM}×{MB}} \\ $$$${since}\:\theta=\mathrm{90}°−\angle{OMB} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{−{l}_{\mathrm{2}} ^{\mathrm{2}} +{OM}^{\mathrm{2}} +{MB}^{\mathrm{2}} }{\mathrm{2}×{OM}×{MB}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{−{l}_{\mathrm{2}} ^{\mathrm{2}} +\frac{\mathrm{2}\left({l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} \right)−{l}^{\mathrm{2}} }{\mathrm{4}}+\left(\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}×\frac{\sqrt{\mathrm{2}\left({l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} \right)−{l}^{\mathrm{2}} }}{\mathrm{2}}×\frac{{l}}{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{−\mathrm{4}{l}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}\left({l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} \right)−{l}^{\mathrm{2}} +{l}^{\mathrm{2}} }{\mathrm{2}{l}\sqrt{\mathrm{2}\left({l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} \right)−{l}^{\mathrm{2}} }} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{{l}_{\mathrm{1}} ^{\mathrm{2}} −{l}_{\mathrm{2}} ^{\mathrm{2}} }{{l}\sqrt{\mathrm{2}\left({l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} \right)−{l}^{\mathrm{2}} }} \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \frac{{l}_{\mathrm{1}} ^{\mathrm{2}} −{l}_{\mathrm{2}} ^{\mathrm{2}} }{{l}\sqrt{\mathrm{2}\left({l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} \right)−{l}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\alpha=\frac{{OA}^{\mathrm{2}} +{OM}^{\mathrm{2}} −{AM}^{\mathrm{2}} }{\mathrm{2}×{OA}×{OM}} \\ $$$$\mathrm{cos}\:\alpha=\frac{{l}_{\mathrm{1}} ^{\mathrm{2}} +\frac{\mathrm{2}\left({l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} \right)−{l}^{\mathrm{2}} }{\mathrm{4}}−\left(\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}×{l}_{\mathrm{1}} ×\frac{\sqrt{\mathrm{2}\left({l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} \right)−{l}^{\mathrm{2}} }}{\mathrm{2}}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{3}{l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} −{l}^{\mathrm{2}} }{\mathrm{2}{l}_{\mathrm{1}} \sqrt{\mathrm{2}\left({l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} \right)−{l}^{\mathrm{2}} }} \\ $$$$\Rightarrow\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{3}{l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} −{l}^{\mathrm{2}} }{\mathrm{2}{l}_{\mathrm{1}} \sqrt{\mathrm{2}\left({l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} \right)−{l}^{\mathrm{2}} }} \\ $$$$\Rightarrow\beta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{3}{l}_{\mathrm{2}} ^{\mathrm{2}} +{l}_{\mathrm{1}} ^{\mathrm{2}} −{l}^{\mathrm{2}} }{\mathrm{2}{l}_{\mathrm{2}} \sqrt{\mathrm{2}\left({l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} \right)−{l}^{\mathrm{2}} }} \\ $$

Commented by ajfour last updated on 23/Jun/18

Brilliant thought sir, but l_1 , l_2 are themselves dependent on α, β ; Sir..

$${Brilliant}\:{thought}\:{sir}, \\ $$$$\:{but}\:\:{l}_{\mathrm{1}} ,\:{l}_{\mathrm{2}} \:{are}\:{themselves} \\ $$$${dependent}\:{on}\:\alpha,\:\beta\:;\:{Sir}.. \\ $$

Commented by MrW3 last updated on 23/Jun/18

yes. the problem is still not easy to solve. one can not get a closed formula for case d≠0 to calculate θ directly.

$${yes}.\: \\ $$$${the}\:{problem}\:{is}\:{still}\:{not}\:{easy}\:{to}\:{solve}. \\ $$$${one}\:{can}\:{not}\:{get}\:{a}\:{closed}\:{formula}\:{for} \\ $$$${case}\:{d}\neq\mathrm{0}\:{to}\:{calculate}\:\theta\:{directly}.\: \\ $$

Commented by ajfour last updated on 23/Jun/18

Thank you anyway Sir.

$${Thank}\:{you}\:{anyway}\:{Sir}. \\ $$

Commented by MrW3 last updated on 24/Jun/18

I don′t think it′s possible to find a closed formula to calculate the angle θ directly. If we know the angle ∠BAC, let′s say ϕ, then we can also find θ. I think we can get a single equation, though a complicated one, for ϕ, something like F(ϕ,a,b,d,l)=0, not in form of ϕ=F(a,b,d,l).

$${I}\:{don}'{t}\:{think}\:{it}'{s}\:{possible}\:{to}\:{find}\:{a} \\ $$$${closed}\:{formula}\:{to}\:{calculate}\:{the}\:{angle} \\ $$$$\theta\:{directly}.\:{If}\:{we}\:{know}\:{the}\:{angle}\:\angle{BAC}, \\ $$$${let}'{s}\:{say}\:\varphi,\:{then}\:{we}\:{can}\:{also}\:{find}\:\theta.\: \\ $$$${I}\:{think}\:{we}\:{can}\:{get}\:{a}\:{single}\:{equation}, \\ $$$${though}\:{a}\:{complicated}\:{one},\:{for}\:\varphi, \\ $$$${something}\:{like}\:{F}\left(\varphi,{a},{b},{d},{l}\right)=\mathrm{0},\:{not} \\ $$$${in}\:{form}\:{of}\:\varphi={F}\left({a},{b},{d},{l}\right). \\ $$

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] ___________________________ .

$$\left\{\:\:{d}_{\mathrm{1}} +{a}\mathrm{cos}\:\alpha=\frac{{l}}{\mathrm{2}}\mathrm{cos}\:\theta\:\:\:\:\right\}×\mathrm{tan}\:\boldsymbol{\alpha} \\ $$$$\:\left\{{d}_{\mathrm{2}} +{b}\mathrm{cos}\:\beta\:=\frac{{l}}{\mathrm{2}}\mathrm{cos}\:\theta\:\:\:\:\right\}×\mathrm{tan}\:\beta \\ $$$$\:\:\:{Since}\:\:\:{d}_{\mathrm{1}} \mathrm{tan}\:\alpha\:=\:{d}_{\mathrm{2}} \mathrm{tan}\:\beta\:=\boldsymbol{{h}} \\ $$$${a}\mathrm{sin}\:\alpha−{b}\mathrm{sin}\:\beta\:=\frac{{l}}{\mathrm{2}}\mathrm{cos}\:\theta\left(\mathrm{tan}\:\alpha−\mathrm{tan}\:\beta\right) \\ $$$${Now}\:\:\:{a}\mathrm{sin}\:\alpha−{b}\mathrm{sin}\:\beta\:=\:{l}\mathrm{sin}\:\theta \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:\:\:\:\mathrm{tan}\:\boldsymbol{\alpha}−\mathrm{tan}\:\boldsymbol{\beta}\:=\:\mathrm{2tan}\:\boldsymbol{\theta} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${Considering}\:{the}\:\bigtriangleup\:{with}\:{sides} \\ $$$${a},\:{b},\:{and}\:\rho\:{we}\:{have}\: \\ $$$$\:\:\:\frac{\rho}{\mathrm{sin}\:\left(\alpha+\beta\right)}=\frac{{a}}{\mathrm{sin}\:\left(\beta+\delta\right)}=\frac{{b}}{\mathrm{sin}\:\left(\alpha−\delta\right)} \\ $$$$\alpha=\mathrm{sin}^{−\mathrm{1}} \left[\frac{{b}\mathrm{sin}\:\left(\alpha+\beta\right)}{\rho}\right]+\delta \\ $$$$\beta=\mathrm{sin}^{−\mathrm{1}} \left[\frac{{a}\mathrm{sin}\:\left(\alpha+\beta\right)}{\rho}\right]−\delta \\ $$$$ \\ $$$${where}\:\left({as}\:{can}\:{easily}\:{be}\:{seen}\right) \\ $$$$\:\:\boldsymbol{\rho}^{\mathrm{2}} =\left(\boldsymbol{{l}}\mathrm{cos}\:\boldsymbol{\theta}−\boldsymbol{{d}}\right)^{\mathrm{2}} +\boldsymbol{{l}}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\theta} \\ $$$$\:\:\:\mathrm{tan}\:\boldsymbol{\delta}\:=\:\frac{\boldsymbol{{l}}\mathrm{sin}\:\boldsymbol{\theta}}{\boldsymbol{{l}}\mathrm{cos}\:\boldsymbol{\theta}−\boldsymbol{{d}}} \\ $$$$\mathrm{cos}\:\left(\boldsymbol{\alpha}+\boldsymbol{\beta}\right)=\frac{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{\rho}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{{ab}}} \\ $$$$\mathrm{sin}\:\left(\boldsymbol{\alpha}+\boldsymbol{\beta}\right)=\sqrt{\mathrm{1}−\left[\frac{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{\rho}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{{ab}}}\right]^{\mathrm{2}} }\: \\ $$$$\rho^{\mathrm{2}} −{b}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \left(\alpha+\beta\right)=\rho^{\mathrm{2}} −{b}^{\mathrm{2}} +\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\rho^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{4}{a}^{\mathrm{2}} \left(\rho^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+\left[{a}^{\mathrm{2}} −\left(\rho^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\right]^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$$\:\:\:\:=\left[\frac{{a}^{\mathrm{2}} +\rho^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{a}}\right]^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{2}{ab}\mathrm{sin}\:\left(\alpha+\beta\right)=\sqrt{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\rho^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow{b}\mathrm{sin}\:\left(\alpha+\beta\right)=\frac{\sqrt{\left[\left({a}+{b}\right)^{\mathrm{2}} −\rho^{\mathrm{2}} \right]\left[\rho^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} \right]}}{\mathrm{2}{a}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\mathrm{2tan}\:\boldsymbol{\theta}\:=\:\mathrm{tan}\:\alpha−\mathrm{tan}\:\beta \\ $$$$\:\:\:\:=\mathrm{tan}\:\left\{\mathrm{sin}^{−\mathrm{1}} \left[\frac{{b}\mathrm{sin}\:\left(\alpha+\beta\right)}{\rho}\right]+\delta\right\} \\ $$$$\:\:\:\:\:\:\:\:−\mathrm{tan}\:\left\{\mathrm{sin}^{−\mathrm{1}} \left[\frac{{a}\mathrm{sin}\:\left(\alpha+\beta\right)}{\rho}\right]−\delta\right\} \\ $$$$\:\:=\mathrm{tan}\:\left\{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}\mathrm{sin}\:\left(\alpha+\beta\right)}{\sqrt{\rho^{\mathrm{2}} −{b}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \left(\alpha+\beta\right)}}\right)+\boldsymbol{\delta}\right\} \\ $$$$\:\:\:\:\:\:−\mathrm{tan}\:\left\{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{a}\mathrm{sin}\:\left(\alpha+\beta\right)}{\sqrt{\rho^{\mathrm{2}} −{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \left(\alpha+\beta\right)}}−\boldsymbol{\delta}\right\}\right. \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\mathrm{2tan}\:\theta\: \\ $$$$\:\:=\mathrm{tan}\:\left[\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\left[\left({a}+{b}\right)^{\mathrm{2}} −\rho^{\mathrm{2}} \right]\left[\rho^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} \right]}}{{a}^{\mathrm{2}} +\rho^{\mathrm{2}} −{b}^{\mathrm{2}} }+\delta\right] \\ $$$$\:\:\:\:−\mathrm{tan}\:\left[\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\left[\left({a}+{b}\right)^{\mathrm{2}} −\rho^{\mathrm{2}} \right]\left[\rho^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} \right]}}{{b}^{\mathrm{2}} +\rho^{\mathrm{2}} −{a}^{\mathrm{2}} }−\delta\right]\: \\ $$$$\:\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\:. \\ $$

Commented by MrW3 last updated on 24/Jun/18

this eqn. is correct sir. I checked different values for a,b,d and l. and the solution is right. wonderful working!

$${this}\:{eqn}.\:{is}\:{correct}\:{sir}.\:{I}\:{checked} \\ $$$${different}\:{values}\:{for}\:{a},{b},{d}\:{and}\:{l}.\:{and} \\ $$$${the}\:{solution}\:{is}\:{right}. \\ $$$${wonderful}\:{working}! \\ $$

Commented by ajfour last updated on 25/Jun/18

Thanks you Sir!

$${Thanks}\:{you}\:{Sir}! \\ $$

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))) .

$$\mathrm{2tan}\:\theta \\ $$$$\:=\:{tan}\:\left\{\mathrm{tan}^{−\mathrm{1}} \left[\frac{\sqrt{\left[\left({a}+{b}\right)^{\mathrm{2}} −\rho^{\mathrm{2}} \right]\left[\rho^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} \right]}}{{a}^{\mathrm{2}} +\rho^{\mathrm{2}} −{b}^{\mathrm{2}} }\right]+\delta\right\} \\ $$$$\:\:\:−{tan}\:\left\{\:\mathrm{tan}^{−\mathrm{1}} \left[\frac{\sqrt{\left[\left({a}+{b}\right)^{\mathrm{2}} −\rho^{\mathrm{2}} \right]\left[\rho^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} \right]}}{{b}^{\mathrm{2}} +\rho^{\mathrm{2}} −{a}^{\mathrm{2}} }\right]−\delta\right\} \\ $$$${where} \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\rho}^{\mathrm{2}} =\:\boldsymbol{{l}}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\theta}+\left(\boldsymbol{{l}}\mathrm{cos}\:\boldsymbol{\theta}−\boldsymbol{{d}}\right)^{\mathrm{2}} \\ $$$${and}\:\:\boldsymbol{\delta}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\boldsymbol{{l}}\mathrm{sin}\:\boldsymbol{\theta}}{\boldsymbol{{l}}\mathrm{cos}\:\boldsymbol{\theta}−\boldsymbol{{d}}}\right)\:. \\ $$

Commented by ajfour last updated on 24/Jun/18

To proceed beyond this , isn′t possible for me. This is all i could arrive at, but ∠BAC or/and ∠ABD is/are eliminated, eventually.

$${To}\:{proceed}\:{beyond}\:{this}\:,\:{isn}'{t} \\ $$$${possible}\:{for}\:{me}.\:{This}\:{is}\:{all}\:{i} \\ $$$${could}\:{arrive}\:{at},\:{but}\:\angle{BAC}\:\:{or}/{and} \\ $$$$\angle{ABD}\:{is}/{are}\:{eliminated},\:{eventually}. \\ $$

Commented by MrW3 last updated on 24/Jun/18

That′s great sir! You have at the end one single equation with only one unknown. The eqn. is even more simple than I ′ve expected.

$${That}'{s}\:{great}\:{sir}!\:{You}\:{have}\:{at}\:{the}\:{end} \\ $$$${one}\:{single}\:{equation}\:{with}\:{only}\:{one} \\ $$$${unknown}.\:{The}\:{eqn}.\:{is}\:{even}\:{more}\:{simple} \\ $$$${than}\:{I}\:'{ve}\:{expected}. \\ $$

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