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Question Number 145282 by puissant last updated on 03/Jul/21

Answered by Olaf_Thorendsen last updated on 03/Jul/21

$$t\mathrm{est}\mathrm{le}\mathrm{temps}\mathrm{en}\mathrm{heures}.$$$$\mathrm{Angle}\mathrm{parcouru}\mathrm{par}\mathrm{la}$$$$\mathrm{petite}\mathrm{aiguille}:$$$${\theta }_p\left(t\right)=\frac{2\pi }{12}(t-12)=\frac{\pi }{6}(t-12)$$$$\mathrm{Angle}\mathrm{parcouru}\mathrm{par}\mathrm{la}$$$$\mathrm{grande}\mathrm{aiguille}:$$$${\theta }_g\left(t\right)=\frac{2\pi }{1}(t-12)=2\pi (t-12)$$$$\mathrm{Superposition}\mathrm{des}\mathrm{aiguilles}\mathrm{quand}$$$$\mathrm{les}\mathrm{deux}\mathrm{angles}\mathrm{sont}\mathrm{egaux}\mathrm{modulo}2\pi :$$$${\theta }_g\left(t\right)={\theta }_p\left(t\right)+2k\pi$$$$2\pi (t-12)=\frac{\pi }{6}(t-12)+2k\pi$$$$(t-12)=\frac{1}{12}(t-12)+k$$$$\frac{11}{12}(t-12)=k$$$$t=\frac{12}{11}k+12$$$$\mathrm{La}\mathrm{periode}\mathrm{de}\mathrm{superposition}\mathrm{est}\mathrm{donc}$$$$\mathrm{T}=\frac{12}{11}\left(\mathrm{en}\mathrm{heures}\right)\left(\mathrm{Q}2\right)$$$$(1\mathrm{heure},5\mathrm{minutes}\mathrm{et}27,27\mathrm{secondes})$$$$\mathrm{Pour}k=3\left(3\mathrm{eme}\mathrm{superposition}\right):$$$$t=\frac{12}{11}\times 3+12=\frac{168}{11}\left(\mathrm{en}\mathrm{heures}\right)/\left(\mathrm{Q}1\right)$$$$(\mathrm{a}15\mathrm{heures},16\mathrm{minutes}\mathrm{et}21,82\mathrm{secondes})$$

Commented by puissant last updated on 03/Jul/21

$$\mathrm{merci}\mathrm{beaucoup}$$

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