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Question Number 84785 by john santu last updated on 16/Mar/20
at what time is the short clock and long hour hand form an angle of 180 degrees?
$$\mathrm{at}\:\mathrm{what}\:\mathrm{time}\:\mathrm{is}\:\mathrm{the}\:\mathrm{short}\:\mathrm{clock}\: \\ $$$$\mathrm{and}\:\mathrm{long}\:\mathrm{hour}\:\mathrm{hand}\:\mathrm{form}\:\mathrm{an}\: \\ $$$$\mathrm{angle}\:\mathrm{of}\:\mathrm{180}\:\mathrm{degrees}? \\ $$
Commented by mr W last updated on 16/Mar/20
hour and minute hand?
$${hour}\:{and}\:{minute}\:{hand}? \\ $$
Commented by john santu last updated on 16/Mar/20
yes .
$$\mathrm{yes}\:. \\ $$
Answered by mr W last updated on 16/Mar/20
at 0:00h both hands are at position θ=0. at time h hour and m minute, the position of minute hand is θ_m =m the position of hour hand is θ_h =(h+(m/(60)))×5=5h+(m/(12)) when both hands form an angle of 180°, it means ∣θ_m −θ_h ∣=30 m−5h−(m/(12))=±30 ((11m)/(12))−5h=±30 ⇒11m−60h=±360 for 11m−60h=360 ⇒m=((60(h+6))/(11))≤60 ⇒h≤5 h=0⇒time 0^h 32^m 43^s h=1⇒time 1^h 38^m 10^s h=2⇒time 2^h 43^m 38^s h=3⇒time 3^h 49^m 05^s h=4⇒time 4^h 54^m 32^s h=5⇒time 5^h 60^m 00^s ⇒6^h 00^m 00^s for 11m−60h=−360 ⇒m=((60(h−6))/(11))≥0 ⇒h≥6 h=6⇒time 6^h 00^m 00^s h=7⇒time 7^h 05^m 27^s h=8⇒time 8^h 10^m 54^s h=9⇒time 9^h 16^m 21^s h=10⇒time 10^h 21^m 49^s h=11⇒time 11^h 27^m 16^s h=12⇒time 12^h 32^m 43^s
$${at}\:\mathrm{0}:\mathrm{00}{h}\:{both}\:{hands}\:{are}\:{at}\:{position} \\ $$$$\theta=\mathrm{0}. \\ $$$$ \\ $$$${at}\:{time}\:{h}\:{hour}\:{and}\:{m}\:{minute},\:{the}\:{position}\:{of} \\ $$$${minute}\:{hand}\:{is} \\ $$$$\theta_{{m}} ={m} \\ $$$${the}\:{position}\:{of}\:{hour}\:{hand}\:{is} \\ $$$$\theta_{{h}} =\left({h}+\frac{{m}}{\mathrm{60}}\right)×\mathrm{5}=\mathrm{5}{h}+\frac{{m}}{\mathrm{12}} \\ $$$$ \\ $$$${when}\:{both}\:{hands}\:{form}\:{an}\:{angle}\:{of} \\ $$$$\mathrm{180}°,\:{it}\:{means} \\ $$$$\mid\theta_{{m}} −\theta_{{h}} \mid=\mathrm{30} \\ $$$${m}−\mathrm{5}{h}−\frac{{m}}{\mathrm{12}}=\pm\mathrm{30} \\ $$$$\frac{\mathrm{11}{m}}{\mathrm{12}}−\mathrm{5}{h}=\pm\mathrm{30} \\ $$$$\Rightarrow\mathrm{11}{m}−\mathrm{60}{h}=\pm\mathrm{360} \\ $$$$ \\ $$$${for}\:\mathrm{11}{m}−\mathrm{60}{h}=\mathrm{360} \\ $$$$\Rightarrow{m}=\frac{\mathrm{60}\left({h}+\mathrm{6}\right)}{\mathrm{11}}\leqslant\mathrm{60}\:\Rightarrow{h}\leqslant\mathrm{5} \\ $$$${h}=\mathrm{0}\Rightarrow{time}\:\mathrm{0}^{{h}} \mathrm{32}^{{m}} \mathrm{43}^{{s}} \\ $$$${h}=\mathrm{1}\Rightarrow{time}\:\mathrm{1}^{{h}} \mathrm{38}^{{m}} \mathrm{10}^{{s}} \\ $$$${h}=\mathrm{2}\Rightarrow{time}\:\mathrm{2}^{{h}} \mathrm{43}^{{m}} \mathrm{38}^{{s}} \\ $$$${h}=\mathrm{3}\Rightarrow{time}\:\mathrm{3}^{{h}} \mathrm{49}^{{m}} \mathrm{05}^{{s}} \\ $$$${h}=\mathrm{4}\Rightarrow{time}\:\mathrm{4}^{{h}} \mathrm{54}^{{m}} \mathrm{32}^{{s}} \\ $$$${h}=\mathrm{5}\Rightarrow{time}\:\mathrm{5}^{{h}} \mathrm{60}^{{m}} \mathrm{00}^{{s}} \Rightarrow\mathrm{6}^{{h}} \mathrm{00}^{{m}} \mathrm{00}^{{s}} \\ $$$$ \\ $$$${for}\:\mathrm{11}{m}−\mathrm{60}{h}=−\mathrm{360} \\ $$$$\Rightarrow{m}=\frac{\mathrm{60}\left({h}−\mathrm{6}\right)}{\mathrm{11}}\geqslant\mathrm{0}\:\Rightarrow{h}\geqslant\mathrm{6} \\ $$$${h}=\mathrm{6}\Rightarrow{time}\:\mathrm{6}^{{h}} \mathrm{00}^{{m}} \mathrm{00}^{{s}} \\ $$$${h}=\mathrm{7}\Rightarrow{time}\:\mathrm{7}^{{h}} \mathrm{05}^{{m}} \mathrm{27}^{{s}} \\ $$$${h}=\mathrm{8}\Rightarrow{time}\:\mathrm{8}^{{h}} \mathrm{10}^{{m}} \mathrm{54}^{{s}} \\ $$$${h}=\mathrm{9}\Rightarrow{time}\:\mathrm{9}^{{h}} \mathrm{16}^{{m}} \mathrm{21}^{{s}} \\ $$$${h}=\mathrm{10}\Rightarrow{time}\:\mathrm{10}^{{h}} \mathrm{21}^{{m}} \mathrm{49}^{{s}} \\ $$$${h}=\mathrm{11}\Rightarrow{time}\:\mathrm{11}^{{h}} \mathrm{27}^{{m}} \mathrm{16}^{{s}} \\ $$$${h}=\mathrm{12}\Rightarrow{time}\:\mathrm{12}^{{h}} \mathrm{32}^{{m}} \mathrm{43}^{{s}} \\ $$
Commented by john santu last updated on 16/Mar/20
waw great sir. i find with my formula θ_h = for hour θ_m = for minute the angle α = ∣30×θ_h −((11)/2)×θ_m ∣^(o ) if < 180^o and β = 180^o −α , if > 180^o
$$\mathrm{waw}\:\mathrm{great}\:\mathrm{sir}. \\ $$$$\mathrm{i}\:\mathrm{find}\:\mathrm{with}\:\mathrm{my}\:\mathrm{formula} \\ $$$$\theta_{\mathrm{h}} \:=\:\mathrm{for}\:\mathrm{hour} \\ $$$$\theta_{\mathrm{m}} \:=\:\mathrm{for}\:\mathrm{minute} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\alpha\:=\:\mid\mathrm{30}×\theta_{\mathrm{h}} −\frac{\mathrm{11}}{\mathrm{2}}×\theta_{\mathrm{m}} \mid^{\mathrm{o}\:} \\ $$$$\mathrm{if}\:<\:\mathrm{180}^{\mathrm{o}} \\ $$$$\mathrm{and}\:\beta\:=\:\mathrm{180}^{\mathrm{o}} −\alpha\:,\:\mathrm{if}\:>\:\mathrm{180}^{\mathrm{o}} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 16/Mar/20
the calculated instants are theoretic, based on the assumption that the hands move continuously. in reality the hands of a classic mechanic clock don′t move continuously. for example, the second hand moves a step every second, the minute hand moves a step every 60 seconds, the hour hand moves a step every 12 minutes. when we consider this effect, we′ll get very different and interesting things. see Q17449 for example.
$${the}\:{calculated}\:{instants}\:{are}\:{theoretic}, \\ $$$${based}\:{on}\:{the}\:{assumption}\:{that}\:{the}\:{hands} \\ $$$${move}\:{continuously}.\:{in}\:{reality}\:{the} \\ $$$${hands}\:{of}\:{a}\:{classic}\:{mechanic}\:{clock} \\ $$$${don}'{t}\:{move}\:{continuously}.\:{for}\:{example}, \\ $$$${the}\:{second}\:{hand}\:{moves}\:{a}\:{step}\:{every} \\ $$$${second},\:{the}\:{minute}\:{hand}\:{moves}\:{a}\:{step} \\ $$$${every}\:\mathrm{60}\:{seconds},\:{the}\:{hour}\:{hand}\:{moves} \\ $$$${a}\:{step}\:{every}\:\mathrm{12}\:{minutes}.\:{when}\:{we}\: \\ $$$${consider}\:{this}\:{effect},\:{we}'{ll}\:{get}\:{very} \\ $$$${different}\:{and}\:{interesting}\:{things}. \\ $$$${see}\:{Q}\mathrm{17449}\:{for}\:{example}. \\ $$

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