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Question Number 154204 by mathdanisur last updated on 15/Sep/21

Commented by Rasheed.Sindhi last updated on 16/Sep/21

$$\mathbf{Sorry}\mathrm{that}\mathrm{I}\mathrm{neglect}‘‘\frac{\tan \alpha }{\tan \beta }\in {\mathbb{Z}}^{+}{}^{″}.$$$$\mathrm{Now}\mathrm{the}\mathrm{question}\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{for}$$$$\mathrm{some}\mathrm{fixed}\mathrm{answers}.$$

Answered by Rasheed.Sindhi last updated on 15/Sep/21

$$\bullet △{\mathrm{OA}}_1\mathrm{C}inwhich\mathrm{\angle }{\mathrm{COA}}_1=\alpha +\beta$$$$\frac{\sin (\alpha +\beta )}{2}=\frac{\sin \mathrm{\angle }{\mathrm{OCA}}_1}{1}$$$$\mathrm{C}={\sin }^{-1}\left(\frac{\sin (\alpha +\beta )}{2}\right)$$$$\mathrm{\angle }{\mathrm{A}}_{\mathrm{n}}{\mathrm{OB}}_2=\alpha +\beta +{\mathrm{OCA}}_1$$$$\mathrm{\angle }{\mathrm{A}}_{\mathrm{n}}{\mathrm{OB}}_2=\alpha +\beta +{\sin }^{-1}\left(\frac{\sin (\alpha +\beta )}{2}\right)$$$$\mathrm{\angle }{\mathrm{OA}}_1\mathrm{C}=\pi -(\alpha +\beta +\mathrm{C})$$$$\mathrm{\angle }{\mathrm{OA}}_1\mathrm{C}=\pi -(\alpha +\beta +{\sin }^{-1}\left(\frac{\sin (\alpha +\beta )}{2}\right))$$$$\bullet △{\mathrm{OA}}_{\mathrm{n}}\mathrm{D}inwhich\mathrm{\angle }{\mathrm{DOA}}_{\mathrm{n}}=\beta$$$$\frac{\sin \beta }{1}=\frac{\mathrm{sinD}}{\mathrm{n}}$$$$\bullet △{\mathrm{OA}}_1\mathrm{E}inwhich\mathrm{\angle }{\mathrm{EOA}}_1=\beta$$$$\frac{\sin \beta }{{\mathrm{A}}_1\mathrm{E}}=\frac{\mathrm{sinE}}{{\mathrm{OA}}_1}$$$$\frac{\sin \beta }{{\mathrm{A}}_1\mathrm{E}}=\frac{\mathrm{sinE}}{1}$$$$\bullet △{\mathrm{OA}}_1\mathrm{E}\sim △{\mathrm{OA}}_{\mathrm{n}}\mathrm{D}$$$$\frac{{\mathrm{OA}}_1}{{\mathrm{OA}}_{\mathrm{n}}}=\frac{{\mathrm{A}}_1\mathrm{E}}{{\mathrm{A}}_{\mathrm{n}}\mathrm{D}}$$$$\frac{1}{\mathrm{n}}=\frac{{\mathrm{A}}_1\mathrm{E}}{1}$$$$\mathrm{n}=\frac{1}{{\mathrm{A}}_1\mathrm{E}}$$$$\mathrm{Continue}$$

Commented by mathdanisur last updated on 15/Sep/21

$$\mathbf{S}\mathrm{er},\mathrm{ans}:\mathrm{n}=8\mathrm{and}{\mathrm{A}}_1{\mathrm{OB}}_1=90°$$

Commented by Rasheed.Sindhi last updated on 15/Sep/21

$$Ithinkthattheanswersdepend$$$$uponthevaluesof\alpha \mathrm{\&}\beta .$$

Commented by mathdanisur last updated on 15/Sep/21

$$\mathrm{Dear}\mathbf{S}\mathrm{er},\mathrm{l}\mathrm{think}$$$$\alpha =45$$$$\beta =\mathrm{atan}\left(1/3\right)\to \mathrm{n}=3$$$${\mathrm{A}}_1{\mathrm{OB}}_1=90\to \alpha =\mathrm{atan}\left(3/2\right)$$$$\beta =\mathrm{atan}\left(1/8\right)\to \mathrm{n}=8$$

Commented by Rasheed.Sindhi last updated on 15/Sep/21

$$\mathcal{T}rytosetdifferentvaluesfor\alpha \mathrm{\&}$$$$\beta {you}^{\prime }llgetdifferentvaluesof$$$${\mathrm{A}}_1{\mathrm{OB}}_1\mathrm{\&}\mathrm{n}(\mathrm{although}\mathrm{may}\mathrm{be}\mathrm{not}$$$$\mathrm{whole}\mathrm{number})$$$$Youmaydrawsuchfigureforn=6$$$$inthatcase\alpha \mathrm{\&}\beta willchange!$$

Commented by Rasheed.Sindhi last updated on 15/Sep/21

$$Yourprovidedfigureisfitforn=4$$$$Nowtrytomeasure\alpha \mathrm{\&}\beta .Forthose$$$$valuesof\alpha \mathrm{\&}\beta ,nwillbecertainly$$$$4and{\mathrm{A}}_1{\mathrm{OB}}_{1}willgetothervalue$$$$than90,asyoucanseefromthe$$$$figure.$$

Commented by mathdanisur last updated on 15/Sep/21

$$\mathrm{Thank}\mathrm{you}\mathbf{S}\mathrm{er}\mathrm{for}\mathrm{solution}$$

Commented by Rasheed.Sindhi last updated on 15/Sep/21

$$\mathcal{BTW}{what}^{\prime }syoursourceofyour$$$$questionsser?$$

Commented by mathdanisur last updated on 15/Sep/21

$$\mathrm{From}\mathrm{the}\mathrm{school}\mathrm{l}\mathrm{went}\mathrm{to}\mathrm{Ser}$$

Commented by mathdanisur last updated on 15/Sep/21

$$\mathrm{and}\mathrm{yes}\mathrm{Ser}\mathrm{n}=4\mathrm{angle}60°$$

Commented by mathdanisur last updated on 15/Sep/21

$$\mathrm{dear}\mathrm{Ser},\mathrm{can}\mathrm{you}\mathrm{contunie}?\mathrm{please}$$

Commented by Rasheed.Sindhi last updated on 16/Sep/21

$${\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{be}\mathrm{happy}\mathrm{if}\mathrm{I}\mathrm{can}\mathrm{answer}\mathrm{any}\mathrm{of}$$$$\mathrm{your}\mathrm{questions}.\mathrm{But}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{an}\mathrm{expert}.$$$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{only}\mathbf{trying}\mathrm{to}\mathrm{solve}\mathrm{very}\mathrm{few}\mathrm{questions}.$$$$\mathrm{Anyway}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{worry}\mathrm{there}\mathrm{are}\mathrm{so}$$$$\mathrm{many}\mathrm{experts}\mathrm{in}\mathrm{the}\mathrm{forum}.\mathrm{Of}\mathrm{course}$$$$\mathrm{this}\mathrm{forum}\mathrm{is}\mathrm{a}\mathrm{good}\mathrm{place}\mathrm{to}\mathrm{learn}...$$$$\mathrm{Again}\mathrm{request}\mathrm{you}\mathrm{to}\mathrm{share}\mathrm{your}$$$$\mathrm{solutions}\mathrm{also}.\mathrm{Thanks}\mathrm{ser}.$$

Commented by mathdanisur last updated on 16/Sep/21

$$\mathrm{THANK}\mathrm{YOU}\mathrm{DEAR}\mathbf{SER}$$

Answered by Rasheed.Sindhi last updated on 17/Sep/21

$$\mathcal{A}n\mathcal{E}asy\mathcal{C}ase:\mathrm{\angle }{A}_{1}O{B}_1=90$$$$△{COA}_1,△{DOA}_{n}arerighttriangles.$$$$\stackrel{△{COA}_1:}{\mathrm{\angle }{COA}_1=\alpha +\beta },\stackrel{△{DOA}_n:}{\mathrm{\angle }{DOA}_n=\beta }$$$$\tan \mathrm{\angle }{COA}_1=\tan (\alpha +\beta )=\frac{2}{1}$$$$\alpha +\beta ={\tan }^{-1}\left(2\right)$$$$\tan \mathrm{\angle }{DOA}_n=\tan \beta =\frac{1}{n}\Rightarrow \beta ={\tan }^{-1}\left(\frac{1}{n}\right)$$$$\frac{\tan \alpha }{\tan \beta }=\frac{\tan ((\alpha +\beta )-\beta )}{\tan \beta }$$$$=\frac{\tan ({\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(\frac{1}{n}\right))}{\tan \left({\tan }^{-1}\left(\frac{1}{n}\right)\right)}$$$$\overline{)\begin{array}{c}\tan (a-b)=\frac{\tan a-\tan b}{1+\tan a.\tan b}\end{array}}$$$$=\frac{\frac{2-\frac{1}{n}}{1+2\left(\frac{1}{n}\right)}}{\frac{1}{n}}=\frac{\frac{2n-1}{n+2}}{1/n}=\frac{n(2n-1)}{n+2}$$$$\frac{\tan \alpha }{\tan \beta }=\frac{n(2n-1)}{n+2}\in {\mathbb{Z}}^{+}\Rightarrow n=3,8$$$$\mathrm{\angle }{A}_{1}O{B}_1=90\{\begin{array}{l}n=3\\ n=8\end{array}$$

Commented by Rasheed.Sindhi last updated on 17/Sep/21

Commented by mathdanisur last updated on 17/Sep/21

$$\mathrm{Very}\mathrm{nice}\mathrm{thank}\mathrm{you}\mathrm{alot}\mathrm{Ser}$$

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