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Question-154204




Question Number 154204 by mathdanisur last updated on 15/Sep/21
Commented by Rasheed.Sindhi last updated on 16/Sep/21
Sorry that I neglect “ ((tanα )/(tanβ )) ∈Z^+  ”.  Now the question can be solved for  some fixed answers.
$$\boldsymbol{\mathrm{Sorry}}\:\mathrm{that}\:\mathrm{I}\:\mathrm{neglect}\:“\:\frac{\mathrm{tan}\alpha\:}{\mathrm{tan}\beta\:}\:\in\mathbb{Z}^{+} \:''. \\ $$$$\mathrm{Now}\:\mathrm{the}\:\mathrm{question}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{for} \\ $$$$\mathrm{some}\:\mathrm{fixed}\:\mathrm{answers}. \\ $$
Answered by Rasheed.Sindhi last updated on 15/Sep/21
 •△OA_1 C in which ∠COA_1 =α+β     ((sin(α+β) )/2)=((sin∠OCA_1  )/1)     C=sin^(−1) (((sin(α+β) )/2))    ∠A_n OB_2 =α+β+OCA_1     ∠A_n OB_2 =α+β+sin^(−1) (((sin(α+β) )/2))  ∠OA_1 C=π−(α+β+C)  ∠OA_1 C=π−(α+β+sin^(−1) (((sin(α+β) )/2)))  •△OA_n D in which ∠DOA_n =β     ((sinβ )/1)=((sinD )/n)  •△OA_1 E in which ∠EOA_1 =β  ((sinβ )/(A_1 E))=((sinE )/(OA_1 ))  ((sinβ )/(A_1 E))=((sinE )/1)  •△OA_1 E∼△OA_n D  ((OA_1 )/(OA_n ))=((A_1 E)/(A_n D))  (1/n)=((A_1 E)/1)  n=(1/(A_1 E))       Continue
$$\:\bullet\bigtriangleup\mathrm{OA}_{\mathrm{1}} \mathrm{C}\:{in}\:{which}\:\angle\mathrm{COA}_{\mathrm{1}} =\alpha+\beta \\ $$$$\:\:\:\frac{\mathrm{sin}\left(\alpha+\beta\right)\:}{\mathrm{2}}=\frac{\mathrm{sin}\angle\mathrm{OCA}_{\mathrm{1}} \:}{\mathrm{1}} \\ $$$$\:\:\:\mathrm{C}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\left(\alpha+\beta\right)\:}{\mathrm{2}}\right) \\ $$$$\:\:\angle\mathrm{A}_{\mathrm{n}} \mathrm{OB}_{\mathrm{2}} =\alpha+\beta+\mathrm{OCA}_{\mathrm{1}} \\ $$$$\:\:\angle\mathrm{A}_{\mathrm{n}} \mathrm{OB}_{\mathrm{2}} =\alpha+\beta+\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\left(\alpha+\beta\right)\:}{\mathrm{2}}\right) \\ $$$$\angle\mathrm{OA}_{\mathrm{1}} \mathrm{C}=\pi−\left(\alpha+\beta+\mathrm{C}\right) \\ $$$$\angle\mathrm{OA}_{\mathrm{1}} \mathrm{C}=\pi−\left(\alpha+\beta+\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\left(\alpha+\beta\right)\:}{\mathrm{2}}\right)\right) \\ $$$$\bullet\bigtriangleup\mathrm{OA}_{\mathrm{n}} \mathrm{D}\:{in}\:{which}\:\angle\mathrm{DOA}_{\mathrm{n}} =\beta \\ $$$$\:\:\:\frac{\mathrm{sin}\beta\:}{\mathrm{1}}=\frac{\mathrm{sinD}\:}{\mathrm{n}} \\ $$$$\bullet\bigtriangleup\mathrm{OA}_{\mathrm{1}} \mathrm{E}\:{in}\:{which}\:\angle\mathrm{EOA}_{\mathrm{1}} =\beta \\ $$$$\frac{\mathrm{sin}\beta\:}{\mathrm{A}_{\mathrm{1}} \mathrm{E}}=\frac{\mathrm{sinE}\:}{\mathrm{OA}_{\mathrm{1}} } \\ $$$$\frac{\mathrm{sin}\beta\:}{\mathrm{A}_{\mathrm{1}} \mathrm{E}}=\frac{\mathrm{sinE}\:}{\mathrm{1}} \\ $$$$\bullet\bigtriangleup\mathrm{OA}_{\mathrm{1}} \mathrm{E}\sim\bigtriangleup\mathrm{OA}_{\mathrm{n}} \mathrm{D} \\ $$$$\frac{\mathrm{OA}_{\mathrm{1}} }{\mathrm{OA}_{\mathrm{n}} }=\frac{\mathrm{A}_{\mathrm{1}} \mathrm{E}}{\mathrm{A}_{\mathrm{n}} \mathrm{D}} \\ $$$$\frac{\mathrm{1}}{\mathrm{n}}=\frac{\mathrm{A}_{\mathrm{1}} \mathrm{E}}{\mathrm{1}} \\ $$$$\mathrm{n}=\frac{\mathrm{1}}{\mathrm{A}_{\mathrm{1}} \mathrm{E}} \\ $$$$\:\:\:\:\:\mathrm{Continue} \\ $$
Commented by mathdanisur last updated on 15/Sep/21
Ser, ans: n=8 and A_1 OB_1 =90°
$$\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{ans}:\:\mathrm{n}=\mathrm{8}\:\mathrm{and}\:\mathrm{A}_{\mathrm{1}} \mathrm{OB}_{\mathrm{1}} =\mathrm{90}° \\ $$
Commented by Rasheed.Sindhi last updated on 15/Sep/21
I think  that the answers depend  upon the values of α & β.
$${I}\:{think}\:\:{that}\:{the}\:{answers}\:{depend} \\ $$$${upon}\:{the}\:{values}\:{of}\:\alpha\:\&\:\beta. \\ $$
Commented by mathdanisur last updated on 15/Sep/21
Dear Ser, l think  α=45  β=atan(1/3) → n=3  A_1 OB_1 =90 → α=atan(3/2)  β=atan(1/8) → n=8
$$\mathrm{Dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{l}\:\mathrm{think} \\ $$$$\alpha=\mathrm{45} \\ $$$$\beta=\mathrm{atan}\left(\mathrm{1}/\mathrm{3}\right)\:\rightarrow\:\mathrm{n}=\mathrm{3} \\ $$$$\mathrm{A}_{\mathrm{1}} \mathrm{OB}_{\mathrm{1}} =\mathrm{90}\:\rightarrow\:\alpha=\mathrm{atan}\left(\mathrm{3}/\mathrm{2}\right) \\ $$$$\beta=\mathrm{atan}\left(\mathrm{1}/\mathrm{8}\right)\:\rightarrow\:\mathrm{n}=\mathrm{8} \\ $$
Commented by Rasheed.Sindhi last updated on 15/Sep/21
Try to set different values for α &  β you′ll get different values of  A_1 OB_1  & n (although may be not  whole number)  You may draw such figure for n=6  in that case α & β will change!
$$\mathcal{T}{ry}\:{to}\:{set}\:{different}\:{values}\:{for}\:\alpha\:\& \\ $$$$\beta\:{you}'{ll}\:{get}\:{different}\:{values}\:{of} \\ $$$$\mathrm{A}_{\mathrm{1}} \mathrm{OB}_{\mathrm{1}} \:\&\:\mathrm{n}\:\left(\mathrm{although}\:\mathrm{may}\:\mathrm{be}\:\mathrm{not}\right. \\ $$$$\left.\mathrm{whole}\:\mathrm{number}\right) \\ $$$${You}\:{may}\:{draw}\:{such}\:{figure}\:{for}\:{n}=\mathrm{6} \\ $$$${in}\:{that}\:{case}\:\alpha\:\&\:\beta\:{will}\:{change}! \\ $$
Commented by Rasheed.Sindhi last updated on 15/Sep/21
Your provided figure is fit for n=4  Now try to measure α & β.For those  values of α & β , n will be certainly  4 and  A_1 OB_1  will get other value   than 90, as you can see from the  figure.
$${Your}\:{provided}\:{figure}\:{is}\:{fit}\:{for}\:{n}=\mathrm{4} \\ $$$${Now}\:{try}\:{to}\:{measure}\:\alpha\:\&\:\beta.{For}\:{those} \\ $$$${values}\:{of}\:\alpha\:\&\:\beta\:,\:{n}\:{will}\:{be}\:{certainly} \\ $$$$\mathrm{4}\:{and}\:\:\mathrm{A}_{\mathrm{1}} \mathrm{OB}_{\mathrm{1}} \:{will}\:{get}\:{other}\:{value}\: \\ $$$${than}\:\mathrm{90},\:{as}\:{you}\:{can}\:{see}\:{from}\:{the} \\ $$$${figure}. \\ $$
Commented by mathdanisur last updated on 15/Sep/21
Thank you Ser for solution
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{for}\:\mathrm{solution} \\ $$
Commented by Rasheed.Sindhi last updated on 15/Sep/21
BTW  what′s your source of your  questions ser?
$$\mathcal{BTW}\:\:{what}'{s}\:{your}\:{source}\:{of}\:{your} \\ $$$${questions}\:{ser}? \\ $$
Commented by mathdanisur last updated on 15/Sep/21
From the school l went to Ser
$$\mathrm{From}\:\mathrm{the}\:\mathrm{school}\:\mathrm{l}\:\mathrm{went}\:\mathrm{to}\:\mathrm{Ser} \\ $$
Commented by mathdanisur last updated on 15/Sep/21
and yes Ser  n=4  angle 60°
$$\mathrm{and}\:\mathrm{yes}\:\mathrm{Ser}\:\:\mathrm{n}=\mathrm{4}\:\:\mathrm{angle}\:\mathrm{60}° \\ $$
Commented by mathdanisur last updated on 15/Sep/21
dear Ser, can you contunie? please
$$\mathrm{dear}\:\mathrm{Ser},\:\mathrm{can}\:\mathrm{you}\:\mathrm{contunie}?\:\mathrm{please} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Sep/21
I′ll be happy if I can answer any of  your questions.But I′m not an expert.  I′m only trying to solve very few questions.  Anyway don′t worry there are so  many experts in the forum.Of course  this forum is a good place to learn...  Again request you to share your  solutions also. Thanks ser.
$$\mathrm{I}'\mathrm{ll}\:\mathrm{be}\:\mathrm{happy}\:\mathrm{if}\:\mathrm{I}\:\mathrm{can}\:\mathrm{answer}\:\mathrm{any}\:\mathrm{of} \\ $$$$\mathrm{your}\:\mathrm{questions}.\mathrm{But}\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{an}\:\mathrm{expert}. \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{only}\:\boldsymbol{\mathrm{trying}}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{very}\:\mathrm{few}\:\mathrm{questions}. \\ $$$$\mathrm{Anyway}\:\mathrm{don}'\mathrm{t}\:\mathrm{worry}\:\mathrm{there}\:\mathrm{are}\:\mathrm{so} \\ $$$$\mathrm{many}\:\mathrm{experts}\:\mathrm{in}\:\mathrm{the}\:\mathrm{forum}.\mathrm{Of}\:\mathrm{course} \\ $$$$\mathrm{this}\:\mathrm{forum}\:\mathrm{is}\:\mathrm{a}\:\mathrm{good}\:\mathrm{place}\:\mathrm{to}\:\mathrm{learn}… \\ $$$$\mathrm{Again}\:\mathrm{request}\:\mathrm{you}\:\mathrm{to}\:\mathrm{share}\:\mathrm{your} \\ $$$$\mathrm{solutions}\:\mathrm{also}.\:\mathrm{Thanks}\:\mathrm{ser}. \\ $$
Commented by mathdanisur last updated on 16/Sep/21
THANK YOU DEAR SER
$$\mathrm{THANK}\:\mathrm{YOU}\:\mathrm{DEAR}\:\boldsymbol{\mathrm{SER}} \\ $$
Answered by Rasheed.Sindhi last updated on 17/Sep/21
An Easy Case:∠A_1 O B_1 =90  △COA_1 ,△DOA_n  are right triangles.  ∠COA_1 =α+β^(△COA_1 :)  ,     ∠DOA_n =β^(△DOA_n :)   tan∠COA_1 =tan(α+β)=(2/1)  α+β=tan^(−1) (2)  tan∠DOA_n = tanβ=(1/n)⇒β=tan^(−1) ((1/n))   ((tanα )/(tanβ))=((tan( (α+β)−β))/(tanβ))  =((tan(tan^(−1) (2)−tan^(−1) ((1/n))))/(tan(tan^(−1) ((1/n)))))   determinant (((tan(a−b)=((tan a−tan b)/(1+tan a.tan b)))))  =(((2−(1/n))/(1+2((1/n))))/(1/n))=(((2n−1)/(n+2))/(1/n))=((n(2n−1))/(n+2))  ((tanα )/(tanβ))=((n(2n−1))/(n+2)) ∈Z^+ ⇒n=3,8  ∠A_1 O B_1 =90  { ((n=3)),((n=8)) :}
$$\mathcal{A}{n}\:\mathcal{E}{asy}\:\mathcal{C}{ase}:\angle{A}_{\mathrm{1}} {O}\:{B}_{\mathrm{1}} =\mathrm{90} \\ $$$$\bigtriangleup{COA}_{\mathrm{1}} ,\bigtriangleup{DOA}_{{n}} \:{are}\:{right}\:{triangles}. \\ $$$$\overset{\bigtriangleup{COA}_{\mathrm{1}} :} {\angle{COA}_{\mathrm{1}} =\alpha+\beta}\:,\:\:\:\:\:\overset{\bigtriangleup{DOA}_{{n}} :} {\angle{DOA}_{{n}} =\beta} \\ $$$$\mathrm{tan}\angle{COA}_{\mathrm{1}} =\mathrm{tan}\left(\alpha+\beta\right)=\frac{\mathrm{2}}{\mathrm{1}} \\ $$$$\alpha+\beta=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right) \\ $$$$\mathrm{tan}\angle{DOA}_{{n}} =\:\mathrm{tan}\beta=\frac{\mathrm{1}}{{n}}\Rightarrow\beta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}}\right)\: \\ $$$$\frac{\mathrm{tan}\alpha\:}{\mathrm{tan}\beta}=\frac{\mathrm{tan}\left(\:\left(\alpha+\beta\right)−\beta\right)}{\mathrm{tan}\beta} \\ $$$$=\frac{\mathrm{tan}\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}}\right)\right)}{\mathrm{tan}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}}\right)\right)} \\ $$$$\begin{array}{|c|}{\mathrm{tan}\left({a}−{b}\right)=\frac{\mathrm{tan}\:{a}−\mathrm{tan}\:{b}}{\mathrm{1}+\mathrm{tan}\:{a}.\mathrm{tan}\:{b}}}\\\hline\end{array} \\ $$$$=\frac{\frac{\mathrm{2}−\frac{\mathrm{1}}{{n}}}{\mathrm{1}+\mathrm{2}\left(\frac{\mathrm{1}}{{n}}\right)}}{\frac{\mathrm{1}}{{n}}}=\frac{\frac{\mathrm{2}{n}−\mathrm{1}}{{n}+\mathrm{2}}}{\mathrm{1}/{n}}=\frac{{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}{{n}+\mathrm{2}} \\ $$$$\frac{\mathrm{tan}\alpha\:}{\mathrm{tan}\beta}=\frac{{n}\left(\mathrm{2}{n}−\mathrm{1}\right)}{{n}+\mathrm{2}}\:\in\mathbb{Z}^{+} \Rightarrow{n}=\mathrm{3},\mathrm{8} \\ $$$$\angle{A}_{\mathrm{1}} {O}\:{B}_{\mathrm{1}} =\mathrm{90}\:\begin{cases}{{n}=\mathrm{3}}\\{{n}=\mathrm{8}}\end{cases} \\ $$
Commented by Rasheed.Sindhi last updated on 17/Sep/21
Commented by mathdanisur last updated on 17/Sep/21
Very nice thank you alot Ser
$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{alot}\:\mathrm{Ser} \\ $$

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