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Question Number 38190 by ajfour last updated on 22/Jun/18

Commented by ajfour last updated on 22/Jun/18

$$\mathit{W}\mathit{h}\mathit{a}\mathit{t}\mathit{i}\mathit{s}\mathit{t}\mathit{h}\mathit{e}\mathit{t}\mathit{i}\mathit{m}\mathit{e}\mathit{i}\mathit{n}\mathit{t}\mathit{h}\mathit{e}$$$$\mathit{o}\mathit{u}\mathit{t}\mathit{e}\mathit{r}\mathit{c}\mathit{l}\mathit{o}\mathit{c}\mathit{k}?$$$$Ignoretheminutehandofthe$$$$biggerclock.Thenitwillbea$$$$correctquestion,ithinknow.$$

Commented by MrW3 last updated on 23/Jun/18

$$let\theta ={\tan }^{-1}2\approx 63.43°$$$$\frac{\sin \alpha }{\sqrt{5}a}=\frac{\sin (\frac{\pi }{2}+\frac{\pi }{2}-\theta )}{8a}$$$$\Rightarrow \sin \alpha =\frac{\sqrt{5}}{8}\sin \theta =\frac{\sqrt{5}}{8}\times \frac{2}{\sqrt{5}}=\frac{1}{4}$$$$\Rightarrow \alpha ={\sin }^{-1}\frac{1}{4}\approx 14.48°$$$$\Rightarrow \frac{5\times 14.48}{30}=2.41min\equiv 2min24sec$$$$$$$$\frac{\sin \beta }{\sqrt{5}a}=\frac{\sin (\theta +\frac{\pi }{6})}{8a}$$$$\Rightarrow \sin \beta =\frac{\sqrt{5}}{8}\sin (\theta +\frac{\pi }{6})=\frac{\sqrt{5}}{8}(\frac{2}{\sqrt{5}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{5}}\times \frac{1}{2})=\frac{2\sqrt{3}+1}{16}$$$$\Rightarrow \beta ={\sin }^{-1}\left[\frac{2\sqrt{3}+1}{16}\right]\approx 16.2°$$$$\gamma =180°-\beta -30°=133.8°$$$$\Rightarrow \frac{133.8}{30}=4.46H\equiv 4H27M36S$$

Answered by ajfour last updated on 23/Jun/18

$$Ignoringtheminutehand,$$$$thetimeinouterclockbyits$$$$hourhand:$$$$let\mathrm{\angle }ofhourhandofbiggerclock$$$$frompostion6o^{\prime }clckbe\theta .$$$$8a\sin \theta =2a+(8a\cos \theta +a)\tan 30°$$$$8(\sin \theta -\frac{\cos \theta }{\sqrt{3}})=2+\frac{1}{\sqrt{3}}$$$$8(\sqrt{3}\sin \theta -\cos \theta )=1+2\sqrt{3}$$$$\sin (\theta -\frac{\pi }{6})=\frac{1+2\sqrt{3}}{16}$$$$\theta =\frac{\pi }{6}+{\sin }^{-1}\left(\frac{1+2\sqrt{3}}{16}\right)$$$$H=6-\frac{6}{\pi }[\frac{\pi }{6}+{\sin }^{-1}\left(\frac{1+2\sqrt{3}}{16}\right)]$$$$H=5-\frac{6}{\pi }{\sin }^{-1}\left(\frac{1+2\sqrt{3}}{16}\right).$$$$\underset{wrongpreviousanswer}{\mathrm{\_}}$$$$\mathit{t}=[4+\frac{6}{\mathit{\pi }}{\sin }^{-1}\left(\frac{1+2\sqrt{3}}{16}\right)]hours.$$

Commented by MrW3 last updated on 23/Jun/18

$$pleasechecksir.shoulditnotbe$$$$\mathit{t}=[5-\frac{6}{\mathit{\pi }}{\sin }^{-1}\left(\frac{1+2\sqrt{3}}{16}\right)]hours?$$

Commented by ajfour last updated on 23/Jun/18

$$YesSir,itisasyousuggest.$$

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