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-

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Question Number 210935 by mr W last updated on 22/Aug/24

$$atwhattimes,ifexist,arethe$$$$anglesbetwenthehourhand,the$$$$minutehandandthesecondhand$$$$ofaclockexactly120°?$$$$assumethatthehandsoftheclock$$$$moveuniformly.$$

Answered by mahdipoor last updated on 22/Aug/24

$$gettimeish:m:s$$$$\mathrm{\angle }h=\frac{360}{12}h+\frac{360}{12\times 60}m+\frac{360}{12\times 60\times 60}s$$$$=30(h+\frac{m}{60}+\frac{s}{3600})=6(5h+\frac{m}{12}+\frac{s}{720})$$$$\mathrm{\angle }m=\frac{360}{60}m+\frac{360}{60\times 60}s=6(m+\frac{s}{60})$$$$\mathrm{\angle }s=6\left(s\right)$$$$\mathrm{\angle }hm=6\mid 5h-\frac{11m}{12}-\frac{11s}{720}\mid =120or240$$$$\Rightarrow 3600h-660m-11s=\pm 14400or\pm 28800$$$$\Rightarrow \dots \Rightarrow answerjust8:0:0and4:0:0$$$$inbothtime\mathrm{\angle }sm=0\Rightarrow never3handmake{120}^{°}$$

Commented by mr W last updated on 23/Aug/24

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Commented by mr W last updated on 23/Aug/24

$$thereshouldbemoreinstantswhen$$$$theanglebetweenhourhandand$$$$minutehandis120°,forexample:$$$$5:5:27.27$$$$6:10:54.55$$$$8:21:49.09$$$$etc.$$

Commented by mahdipoor last updated on 23/Aug/24

$$igeth,m,s\in W$$

Commented by A5T last updated on 23/Aug/24

$$Letthetimebea:bwhereaisthehourhandand$$$$btheminutehand.$$$$Thenanglebetweenthehourandminutehand$$$$=\mid (a\times 30°+\frac{b}{60}\times 30°)-\frac{360°b}{60}\mid$$$$=\mid 30a+\frac{b}{2}-6b\mid =\mid 30a-\frac{11b}{2}\mid =120°$$$$or360°-\mid 30a-\frac{11b}{2}\mid =120°$$

Answered by mr W last updated on 24/Aug/24

$${let}^{\prime }slookattheinstantxseconds$$$$after12:00.$$$$x\in R,0⩽x<43200$$$${it}^{\prime }sthetimeh:m:s$$$$h=\lfloor \frac{x}{3600}\rfloor$$$$m=\lfloor \frac{x}{60}\rfloor -60\lfloor \frac{x}{3600}\rfloor$$$$s=x-60\lfloor \frac{x}{60}\rfloor$$$$thepositions\left(in°\right)ofthehandsare:$$$${\theta }_s=6s$$$${\theta }_m=6(m+\frac{s}{60})$$$${\theta }_h=30(h+\frac{m}{60}+\frac{s}{3600})$$$$$$$$anglebetweenhourhandandminutehand:$$$$\mathrm{\Delta }{\theta }_{hm}=30(h+\frac{m}{60}+\frac{s}{3600})-6(m+\frac{s}{60})$$$$\mathrm{\Delta }{\theta }_{hm}=30h-\frac{11m}{2}-\frac{11s}{120}$$$$\mathrm{\Delta }{\theta }_{hm}=30\lfloor \frac{x}{3600}\rfloor -\frac{11}{2}(\lfloor \frac{x}{60}\rfloor -60\lfloor \frac{x}{3600}\rfloor )-\frac{11}{120}(x-60\lfloor \frac{x}{60}\rfloor )$$$$\mathrm{\Delta }{\theta }_{hm}=360\lfloor \frac{x}{3600}\rfloor -\frac{11x}{120}$$$$suchthatthisangleis120°,$$$$360\times \lfloor \frac{x}{3600}\rfloor -\frac{11x}{120}=\pm 120,\pm 240$$$$sayx=3600h+kwith0⩽h⩽11,0⩽k<3600$$$$360h-\frac{11(3600h+k)}{120}=\pm 120,\pm 240$$$$\Rightarrow k=\frac{120}{11}(30h\mp 120,\mp 240)$$$$therearefollowingsolutions:$$$$\overline{)\begin{array}{ccccc}h╲\mathrm{\Delta }{\theta }_{hm}& +120& -120& +240& -240\\ 0:& -& 21:\frac{540}{11}& -& 43:\frac{420}{11}\\ 1:& -& 27:\frac{180}{11}& -& 49:\frac{60}{11}\\ 2:& -& 32:\frac{480}{11}& -& 54:\frac{360}{11}\\ 3:& -& 38:\frac{120}{11}& -& -\\ 4:& 0:0& 43:\frac{420}{11}& -& -\\ 5:& 5:\frac{300}{11}& 49:\frac{60}{11}& -& -\\ 6:& 10:\frac{600}{11}& 54:\frac{360}{11}& -& -\\ 7:& 16:\frac{240}{11}& -& -& -\\ 8:& 21:\frac{540}{11}& -& 0:0& -\\ 9:& 27:\frac{180}{11}& -& 5:\frac{300}{11}& -\\ 10:& 32:\frac{480}{11}& -& 10:\frac{600}{11}& -\\ 11:& 38:\frac{120}{11}& -& 16:\frac{240}{11}& -\end{array}}$$$$thatmeansthereare22instantsat$$$$whichtheanglebetweenhourhand$$$$andminutehandis120°.$$$$$$$$anglebetweenminutehandandsecondhand:$$$$\mathrm{\Delta }{\theta }_{ms}=6(m+\frac{s}{60})-6s$$$$\mathrm{\Delta }{\theta }_{ms}=6m-\frac{59s}{10}$$$$\mathrm{\Delta }{\theta }_{ms}=6(\lfloor \frac{x}{60}\rfloor -60\lfloor \frac{x}{3600}\rfloor )-\frac{59}{10}(x-60\lfloor \frac{x}{60}\rfloor )$$$$\mathrm{\Delta }{\theta }_{ms}=-360\lfloor \frac{x}{3600}\rfloor +360\lfloor \frac{x}{60}\rfloor -\frac{59x}{10}$$$$-360\lfloor \frac{x}{3600}\rfloor +360\lfloor \frac{x}{60}\rfloor -\frac{59x}{10}=\pm 120,\pm 240$$$$sayx=3600h+60m+k$$$$with0⩽h⩽11,0⩽m⩽59,0⩽k<60$$$$-360h+360(60h+m)-\frac{59(3600h+60m+k)}{10}=\pm 120,\pm 240$$$$6m-\frac{59k}{10}=\pm 120,\pm 240$$$$for6m-\frac{59k}{10}=120:$$$$60m-1200=59k⩾0\Rightarrow m⩾20$$$$60m-1200=59k<59\times 60\Rightarrow m<79$$$$\Rightarrow 20⩽m⩽59$$$$\Rightarrow k=\frac{10}{59}(6m-120)$$$$for6m-\frac{59k}{10}=-120:$$$$60m+1200=59k⩾0\Rightarrow m⩾-20$$$$60m+1200=59k<59\times 60\Rightarrow m<39$$$$\Rightarrow 0⩽m⩽38$$$$\Rightarrow k=\frac{10}{59}(6m+120)$$$$for6m-\frac{59k}{10}=240:$$$$60m-2400=59k⩾0\Rightarrow m⩾40$$$$60m-2400=59k<59\times 60\Rightarrow m<99$$$$\Rightarrow 40⩽m⩽59$$$$\Rightarrow k=\frac{10}{59}(6m-240)$$$$for6m-\frac{59k}{10}=-240:$$$$60m+2400=59k⩾0\Rightarrow m⩾-40$$$$60m+2400=59k<59\times 60\Rightarrow m<19$$$$\Rightarrow 0⩽m⩽18$$$$\Rightarrow k=\frac{10}{59}(6m+240)$$$$thatmeanstherearetotally$$$$40+39+20+19=118instants$$$$atwhichtheanglebetweenminute$$$$handandsecondhandis120°.$$

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