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Question Number 210935 by mr W last updated on 22/Aug/24
at what times, if exist, are the angles betwen the hour hand, the minute hand and the second hand of a clock exactly 120°? assume that the hands of the clock move uniformly.
$${at}\:{what}\:{times},\:{if}\:{exist},\:{are}\:{the}\: \\ $$$${angles}\:{betwen}\:{the}\:{hour}\:{hand},\:{the} \\ $$$${minute}\:{hand}\:{and}\:{the}\:{second}\:{hand} \\ $$$${of}\:{a}\:{clock}\:{exactly}\:\mathrm{120}°? \\ $$$${assume}\:{that}\:{the}\:{hands}\:{of}\:{the}\:{clock} \\ $$$${move}\:{uniformly}. \\ $$
Answered by mahdipoor last updated on 22/Aug/24
get time is h : m : s ∠h=((360)/(12))h+((360)/(12×60))m+((360)/(12×60×60))s =30(h+(m/(60))+(s/(3600)))=6(5h+(m/(12))+(s/(720))) ∠m=((360)/(60))m+((360)/(60×60))s=6(m+(s/(60))) ∠s=6(s) ∠hm=6∣5h−((11m)/(12))−((11s)/(720))∣=120 or 240 ⇒3600h−660m−11s=±14400 or ±28800 ⇒...⇒ answer just 8:0:0 and 4:0:0 in both time ∠sm=0 ⇒ never 3 hand make 120^°
$${get}\:{time}\:{is}\:\:{h}\::\:{m}\::\:{s} \\ $$$$\angle{h}=\frac{\mathrm{360}}{\mathrm{12}}{h}+\frac{\mathrm{360}}{\mathrm{12}×\mathrm{60}}{m}+\frac{\mathrm{360}}{\mathrm{12}×\mathrm{60}×\mathrm{60}}{s} \\ $$$$=\mathrm{30}\left({h}+\frac{{m}}{\mathrm{60}}+\frac{{s}}{\mathrm{3600}}\right)=\mathrm{6}\left(\mathrm{5}{h}+\frac{{m}}{\mathrm{12}}+\frac{{s}}{\mathrm{720}}\right) \\ $$$$\angle{m}=\frac{\mathrm{360}}{\mathrm{60}}{m}+\frac{\mathrm{360}}{\mathrm{60}×\mathrm{60}}{s}=\mathrm{6}\left({m}+\frac{{s}}{\mathrm{60}}\right) \\ $$$$\angle{s}=\mathrm{6}\left({s}\right) \\ $$$$\angle{hm}=\mathrm{6}\mid\mathrm{5}{h}−\frac{\mathrm{11}{m}}{\mathrm{12}}−\frac{\mathrm{11}{s}}{\mathrm{720}}\mid=\mathrm{120}\:{or}\:\mathrm{240}\: \\ $$$$\Rightarrow\mathrm{3600}{h}−\mathrm{660}{m}−\mathrm{11}{s}=\pm\mathrm{14400}\:{or}\:\pm\mathrm{28800} \\ $$$$\Rightarrow…\Rightarrow\:{answer}\:{just}\:\:\mathrm{8}:\mathrm{0}:\mathrm{0}\:\:{and}\:\mathrm{4}:\mathrm{0}:\mathrm{0}\: \\ $$$${in}\:{both}\:{time}\:\angle{sm}=\mathrm{0}\:\Rightarrow\:{never}\:\mathrm{3}\:{hand}\:{make}\:\mathrm{120}^{°} \\ $$
Commented by mr W last updated on 23/Aug/24
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Commented by mr W last updated on 23/Aug/24
there should be more instants when the angle between hour hand and minute hand is 120°, for example: 5:5:27.27 6:10:54.55 8:21:49.09 etc.
$${there}\:{should}\:{be}\:{more}\:{instants}\:{when} \\ $$$${the}\:{angle}\:{between}\:{hour}\:{hand}\:{and} \\ $$$${minute}\:{hand}\:{is}\:\mathrm{120}°,\:{for}\:{example}: \\ $$$$\mathrm{5}:\mathrm{5}:\mathrm{27}.\mathrm{27} \\ $$$$\mathrm{6}:\mathrm{10}:\mathrm{54}.\mathrm{55} \\ $$$$\mathrm{8}:\mathrm{21}:\mathrm{49}.\mathrm{09} \\ $$$${etc}. \\ $$
Commented by mahdipoor last updated on 23/Aug/24
i get h,m,s∈W
$${i}\:{get}\:\:{h},{m},{s}\in{W} \\ $$
Commented by A5T last updated on 23/Aug/24
Let the time be a:b where a is the hour hand and b the minute hand. Then angle between the hour and minute hand =∣(a×30°+(b/(60))×30°)−((360°b)/(60))∣ =∣30a+(b/2)−6b∣=∣30a−((11b)/2)∣=120° or 360°−∣30a−((11b)/2)∣=120°
$${Let}\:{the}\:{time}\:{be}\:{a}:{b}\:{where}\:{a}\:{is}\:{the}\:{hour}\:{hand}\:{and}\: \\ $$$${b}\:{the}\:{minute}\:{hand}. \\ $$$${Then}\:{angle}\:{between}\:{the}\:{hour}\:{and}\:{minute}\:{hand}\: \\ $$$$=\mid\left({a}×\mathrm{30}°+\frac{{b}}{\mathrm{60}}×\mathrm{30}°\right)−\frac{\mathrm{360}°{b}}{\mathrm{60}}\mid \\ $$$$=\mid\mathrm{30}{a}+\frac{{b}}{\mathrm{2}}−\mathrm{6}{b}\mid=\mid\mathrm{30}{a}−\frac{\mathrm{11}{b}}{\mathrm{2}}\mid=\mathrm{120}° \\ $$$${or}\:\mathrm{360}°−\mid\mathrm{30}{a}−\frac{\mathrm{11}{b}}{\mathrm{2}}\mid=\mathrm{120}° \\ $$
Answered by mr W last updated on 24/Aug/24
let′s look at the instant x seconds after 12:00. x∈R, 0≤x<43200 it′s the time h:m:s h=⌊(x/(3600))⌋ m=⌊(x/(60))⌋−60⌊(x/(3600))⌋ s=x−60⌊(x/(60))⌋ the positions (in °) of the hands are: θ_s =6s θ_m =6(m+(s/(60))) θ_h =30(h+(m/(60))+(s/(3600))) angle between hour hand and minute hand: Δθ_(hm) =30(h+(m/(60))+(s/(3600)))−6(m+(s/(60))) Δθ_(hm) =30h−((11m)/2)−((11s)/(120)) Δθ_(hm) =30⌊(x/(3600))⌋−((11)/2)(⌊(x/(60))⌋−60⌊(x/(3600))⌋)−((11)/(120))(x−60⌊(x/(60))⌋) Δθ_(hm) =360⌊(x/(3600))⌋−((11x)/(120)) such that this angle is 120°, 360×⌊(x/(3600))⌋−((11x)/(120))=±120, ±240 say x=3600h+k with 0≤h≤11, 0≤k<3600 360h−((11(3600h+k))/(120))=±120, ±240 ⇒k=((120)/(11))(30h∓120, ∓240) there are following solutions: determinant (((h╲Δθ_(hm) ),(+120),(−120),(+240),(−240)),((0:),−,(21:((540)/(11))),−,(43:((420)/(11)))),((1:),−,(27:((180)/(11))),−,(49:((60)/(11)))),((2:),−,(32:((480)/(11))),−,(54:((360)/(11)))),((3:),−,(38:((120)/(11))),−,−),((4:),(0:0),(43:((420)/(11))),−,−),((5:),(5:((300)/(11))),(49:((60)/(11))),−,−),((6:),(10:((600)/(11))),(54:((360)/(11))),−,−),((7:),(16:((240)/(11))),−,−,−),((8:),(21:((540)/(11))),−,(0:0),−),((9:),(27:((180)/(11))),−,(5:((300)/(11))),−),((10:),(32:((480)/(11))),−,(10:((600)/(11))),−),((11:),(38:((120)/(11))),−,(16:((240)/(11))),−)) that means there are 22 instants at which the angle between hour hand and minute hand is 120°. angle between minute hand and second hand: Δθ_(ms) =6(m+(s/(60)))−6s Δθ_(ms) =6m−((59s)/(10)) Δθ_(ms) =6(⌊(x/(60))⌋−60⌊(x/(3600))⌋)−((59)/(10))(x−60⌊(x/(60))⌋) Δθ_(ms) =−360⌊(x/(3600))⌋+360⌊(x/(60))⌋−((59x)/(10)) −360⌊(x/(3600))⌋+360⌊(x/(60))⌋−((59x)/(10))=±120, ±240 say x=3600h+60m+k with 0≤h≤11, 0≤m≤59, 0≤k<60 −360h+360(60h+m)−((59(3600h+60m+k))/(10))=±120, ±240 6m−((59k)/(10))=±120, ±240 for 6m−((59k)/(10))=120: 60m−1200=59k≥0 ⇒m≥20 60m−1200=59k<59×60⇒m<79 ⇒20≤m≤59 ⇒k=((10)/(59))(6m−120) for 6m−((59k)/(10))=−120: 60m+1200=59k≥0 ⇒m≥−20 60m+1200=59k<59×60 ⇒m<39 ⇒0≤m≤38 ⇒k=((10)/(59))(6m+120) for 6m−((59k)/(10))=240: 60m−2400=59k≥0 ⇒m≥40 60m−2400=59k<59×60⇒m<99 ⇒40≤m≤59 ⇒k=((10)/(59))(6m−240) for 6m−((59k)/(10))=−240: 60m+2400=59k≥0 ⇒m≥−40 60m+2400=59k<59×60 ⇒m<19 ⇒0≤m≤18 ⇒k=((10)/(59))(6m+240) that means there are totally 40+39+20+19=118 instants at which the angle between minute hand and second hand is 120°.
$${let}'{s}\:{look}\:{at}\:{the}\:{instant}\:{x}\:{seconds} \\ $$$${after}\:\mathrm{12}:\mathrm{00}.\: \\ $$$${x}\in{R},\:\mathrm{0}\leqslant{x}<\mathrm{43200} \\ $$$${it}'{s}\:{the}\:{time}\:{h}:{m}:{s} \\ $$$${h}=\lfloor\frac{{x}}{\mathrm{3600}}\rfloor \\ $$$${m}=\lfloor\frac{{x}}{\mathrm{60}}\rfloor−\mathrm{60}\lfloor\frac{{x}}{\mathrm{3600}}\rfloor \\ $$$${s}={x}−\mathrm{60}\lfloor\frac{{x}}{\mathrm{60}}\rfloor \\ $$$${the}\:{positions}\:\left({in}\:°\right)\:{of}\:{the}\:{hands}\:{are}: \\ $$$$\theta_{{s}} =\mathrm{6}{s} \\ $$$$\theta_{{m}} =\mathrm{6}\left({m}+\frac{{s}}{\mathrm{60}}\right) \\ $$$$\theta_{{h}} =\mathrm{30}\left({h}+\frac{{m}}{\mathrm{60}}+\frac{{s}}{\mathrm{3600}}\right) \\ $$$$ \\ $$$${angle}\:{between}\:{hour}\:{hand}\:{and}\:{minute}\:{hand}: \\ $$$$\Delta\theta_{{hm}} =\mathrm{30}\left({h}+\frac{{m}}{\mathrm{60}}+\frac{{s}}{\mathrm{3600}}\right)−\mathrm{6}\left({m}+\frac{{s}}{\mathrm{60}}\right) \\ $$$$\Delta\theta_{{hm}} =\mathrm{30}{h}−\frac{\mathrm{11}{m}}{\mathrm{2}}−\frac{\mathrm{11}{s}}{\mathrm{120}} \\ $$$$\Delta\theta_{{hm}} =\mathrm{30}\lfloor\frac{{x}}{\mathrm{3600}}\rfloor−\frac{\mathrm{11}}{\mathrm{2}}\left(\lfloor\frac{{x}}{\mathrm{60}}\rfloor−\mathrm{60}\lfloor\frac{{x}}{\mathrm{3600}}\rfloor\right)−\frac{\mathrm{11}}{\mathrm{120}}\left({x}−\mathrm{60}\lfloor\frac{{x}}{\mathrm{60}}\rfloor\right) \\ $$$$\Delta\theta_{{hm}} =\mathrm{360}\lfloor\frac{{x}}{\mathrm{3600}}\rfloor−\frac{\mathrm{11}{x}}{\mathrm{120}} \\ $$$${such}\:{that}\:{this}\:{angle}\:{is}\:\mathrm{120}°, \\ $$$$\mathrm{360}×\lfloor\frac{{x}}{\mathrm{3600}}\rfloor−\frac{\mathrm{11}{x}}{\mathrm{120}}=\pm\mathrm{120},\:\pm\mathrm{240} \\ $$$${say}\:{x}=\mathrm{3600}{h}+{k}\:{with}\:\mathrm{0}\leqslant{h}\leqslant\mathrm{11},\:\mathrm{0}\leqslant{k}<\mathrm{3600} \\ $$$$\mathrm{360}{h}−\frac{\mathrm{11}\left(\mathrm{3600}{h}+{k}\right)}{\mathrm{120}}=\pm\mathrm{120},\:\pm\mathrm{240} \\ $$$$\Rightarrow{k}=\frac{\mathrm{120}}{\mathrm{11}}\left(\mathrm{30}{h}\mp\mathrm{120},\:\mp\mathrm{240}\right) \\ $$$${there}\:{are}\:{following}\:{solutions}: \\ $$$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}{{h}\diagdown\Delta\theta_{{hm}} }&\hline{+\mathrm{120}}&\hline{−\mathrm{120}}&\hline{+\mathrm{240}}&\hline{−\mathrm{240}}\\{\mathrm{0}:}&\hline{−}&\hline{\mathrm{21}:\frac{\mathrm{540}}{\mathrm{11}}}&\hline{−}&\hline{\mathrm{43}:\frac{\mathrm{420}}{\mathrm{11}}}\\{\mathrm{1}:}&\hline{−}&\hline{\mathrm{27}:\frac{\mathrm{180}}{\mathrm{11}}}&\hline{−}&\hline{\mathrm{49}:\frac{\mathrm{60}}{\mathrm{11}}}\\{\mathrm{2}:}&\hline{−}&\hline{\mathrm{32}:\frac{\mathrm{480}}{\mathrm{11}}}&\hline{−}&\hline{\mathrm{54}:\frac{\mathrm{360}}{\mathrm{11}}}\\{\mathrm{3}:}&\hline{−}&\hline{\mathrm{38}:\frac{\mathrm{120}}{\mathrm{11}}}&\hline{−}&\hline{−}\\{\mathrm{4}:}&\hline{\mathrm{0}:\mathrm{0}}&\hline{\mathrm{43}:\frac{\mathrm{420}}{\mathrm{11}}}&\hline{−}&\hline{−}\\{\mathrm{5}:}&\hline{\mathrm{5}:\frac{\mathrm{300}}{\mathrm{11}}}&\hline{\mathrm{49}:\frac{\mathrm{60}}{\mathrm{11}}}&\hline{−}&\hline{−}\\{\mathrm{6}:}&\hline{\mathrm{10}:\frac{\mathrm{600}}{\mathrm{11}}}&\hline{\mathrm{54}:\frac{\mathrm{360}}{\mathrm{11}}}&\hline{−}&\hline{−}\\{\mathrm{7}:}&\hline{\mathrm{16}:\frac{\mathrm{240}}{\mathrm{11}}}&\hline{−}&\hline{−}&\hline{−}\\{\mathrm{8}:}&\hline{\mathrm{21}:\frac{\mathrm{540}}{\mathrm{11}}}&\hline{−}&\hline{\mathrm{0}:\mathrm{0}}&\hline{−}\\{\mathrm{9}:}&\hline{\mathrm{27}:\frac{\mathrm{180}}{\mathrm{11}}}&\hline{−}&\hline{\mathrm{5}:\frac{\mathrm{300}}{\mathrm{11}}}&\hline{−}\\{\mathrm{10}:}&\hline{\mathrm{32}:\frac{\mathrm{480}}{\mathrm{11}}}&\hline{−}&\hline{\mathrm{10}:\frac{\mathrm{600}}{\mathrm{11}}}&\hline{−}\\{\mathrm{11}:}&\hline{\mathrm{38}:\frac{\mathrm{120}}{\mathrm{11}}}&\hline{−}&\hline{\mathrm{16}:\frac{\mathrm{240}}{\mathrm{11}}}&\hline{−}\\\hline\end{array} \\ $$$${that}\:{means}\:{there}\:{are}\:\mathrm{22}\:{instants}\:{at} \\ $$$${which}\:{the}\:{angle}\:{between}\:{hour}\:{hand} \\ $$$${and}\:{minute}\:{hand}\:{is}\:\mathrm{120}°. \\ $$$$ \\ $$$${angle}\:{between}\:{minute}\:{hand}\:{and}\:{second}\:{hand}: \\ $$$$\Delta\theta_{{ms}} =\mathrm{6}\left({m}+\frac{{s}}{\mathrm{60}}\right)−\mathrm{6}{s} \\ $$$$\Delta\theta_{{ms}} =\mathrm{6}{m}−\frac{\mathrm{59}{s}}{\mathrm{10}} \\ $$$$\Delta\theta_{{ms}} =\mathrm{6}\left(\lfloor\frac{{x}}{\mathrm{60}}\rfloor−\mathrm{60}\lfloor\frac{{x}}{\mathrm{3600}}\rfloor\right)−\frac{\mathrm{59}}{\mathrm{10}}\left({x}−\mathrm{60}\lfloor\frac{{x}}{\mathrm{60}}\rfloor\right) \\ $$$$\Delta\theta_{{ms}} =−\mathrm{360}\lfloor\frac{{x}}{\mathrm{3600}}\rfloor+\mathrm{360}\lfloor\frac{{x}}{\mathrm{60}}\rfloor−\frac{\mathrm{59}{x}}{\mathrm{10}} \\ $$$$−\mathrm{360}\lfloor\frac{{x}}{\mathrm{3600}}\rfloor+\mathrm{360}\lfloor\frac{{x}}{\mathrm{60}}\rfloor−\frac{\mathrm{59}{x}}{\mathrm{10}}=\pm\mathrm{120},\:\pm\mathrm{240} \\ $$$${say}\:{x}=\mathrm{3600}{h}+\mathrm{60}{m}+{k}\: \\ $$$${with}\:\mathrm{0}\leqslant{h}\leqslant\mathrm{11},\:\mathrm{0}\leqslant{m}\leqslant\mathrm{59},\:\mathrm{0}\leqslant{k}<\mathrm{60} \\ $$$$−\mathrm{360}{h}+\mathrm{360}\left(\mathrm{60}{h}+{m}\right)−\frac{\mathrm{59}\left(\mathrm{3600}{h}+\mathrm{60}{m}+{k}\right)}{\mathrm{10}}=\pm\mathrm{120},\:\pm\mathrm{240} \\ $$$$\mathrm{6}{m}−\frac{\mathrm{59}{k}}{\mathrm{10}}=\pm\mathrm{120},\:\pm\mathrm{240} \\ $$$${for}\:\mathrm{6}{m}−\frac{\mathrm{59}{k}}{\mathrm{10}}=\mathrm{120}: \\ $$$$\mathrm{60}{m}−\mathrm{1200}=\mathrm{59}{k}\geqslant\mathrm{0}\:\Rightarrow{m}\geqslant\mathrm{20} \\ $$$$\mathrm{60}{m}−\mathrm{1200}=\mathrm{59}{k}<\mathrm{59}×\mathrm{60}\Rightarrow{m}<\mathrm{79} \\ $$$$\Rightarrow\mathrm{20}\leqslant{m}\leqslant\mathrm{59} \\ $$$$\Rightarrow{k}=\frac{\mathrm{10}}{\mathrm{59}}\left(\mathrm{6}{m}−\mathrm{120}\right) \\ $$$${for}\:\mathrm{6}{m}−\frac{\mathrm{59}{k}}{\mathrm{10}}=−\mathrm{120}: \\ $$$$\mathrm{60}{m}+\mathrm{1200}=\mathrm{59}{k}\geqslant\mathrm{0}\:\Rightarrow{m}\geqslant−\mathrm{20} \\ $$$$\mathrm{60}{m}+\mathrm{1200}=\mathrm{59}{k}<\mathrm{59}×\mathrm{60}\:\Rightarrow{m}<\mathrm{39} \\ $$$$\Rightarrow\mathrm{0}\leqslant{m}\leqslant\mathrm{38} \\ $$$$\Rightarrow{k}=\frac{\mathrm{10}}{\mathrm{59}}\left(\mathrm{6}{m}+\mathrm{120}\right) \\ $$$${for}\:\mathrm{6}{m}−\frac{\mathrm{59}{k}}{\mathrm{10}}=\mathrm{240}: \\ $$$$\mathrm{60}{m}−\mathrm{2400}=\mathrm{59}{k}\geqslant\mathrm{0}\:\Rightarrow{m}\geqslant\mathrm{40} \\ $$$$\mathrm{60}{m}−\mathrm{2400}=\mathrm{59}{k}<\mathrm{59}×\mathrm{60}\Rightarrow{m}<\mathrm{99} \\ $$$$\Rightarrow\mathrm{40}\leqslant{m}\leqslant\mathrm{59} \\ $$$$\Rightarrow{k}=\frac{\mathrm{10}}{\mathrm{59}}\left(\mathrm{6}{m}−\mathrm{240}\right) \\ $$$${for}\:\mathrm{6}{m}−\frac{\mathrm{59}{k}}{\mathrm{10}}=−\mathrm{240}: \\ $$$$\mathrm{60}{m}+\mathrm{2400}=\mathrm{59}{k}\geqslant\mathrm{0}\:\Rightarrow{m}\geqslant−\mathrm{40} \\ $$$$\mathrm{60}{m}+\mathrm{2400}=\mathrm{59}{k}<\mathrm{59}×\mathrm{60}\:\Rightarrow{m}<\mathrm{19} \\ $$$$\Rightarrow\mathrm{0}\leqslant{m}\leqslant\mathrm{18} \\ $$$$\Rightarrow{k}=\frac{\mathrm{10}}{\mathrm{59}}\left(\mathrm{6}{m}+\mathrm{240}\right) \\ $$$${that}\:{means}\:{there}\:{are}\:{totally} \\ $$$$\mathrm{40}+\mathrm{39}+\mathrm{20}+\mathrm{19}=\mathrm{118}\:{instants} \\ $$$${at}\:{which}\:{the}\:{angle}\:{between}\:{minute} \\ $$$${hand}\:{and}\:{second}\:{hand}\:{is}\:\mathrm{120}°. \\ $$

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