Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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.

atwhattimes,ifexist,aretheanglesbetwenthehourhand,theminutehandandthesecondhandofaclockexactly120°?assumethatthehandsoftheclockmoveuniformly.

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^°

gettimeish:m:sh=36012h+36012×60m+36012×60×60s=30(h+m60+s3600)=6(5h+m12+s720)m=36060m+36060×60s=6(m+s60)s=6(s)hm=65h11m1211s720∣=120or2403600h660m11s=±14400or±28800...answerjust8:0:0and4:0:0inbothtimesm=0never3handmake120°

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

��

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.

thereshouldbemoreinstantswhentheanglebetweenhourhandandminutehandis120°,forexample:5:5:27.276:10:54.558:21:49.09etc.

Commented by mahdipoor last updated on 23/Aug/24

i get  h,m,s∈W

igeth,m,sW

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°

Letthetimebea:bwhereaisthehourhandandbtheminutehand.Thenanglebetweenthehourandminutehand=∣(a×30°+b60×30°)360°b60=∣30a+b26b∣=∣30a11b2∣=120°or360°30a11b2∣=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°.

letslookattheinstantxsecondsafter12:00.xR,0x<43200itsthetimeh:m:sh=x3600m=x6060x3600s=x60x60thepositions(in°)ofthehandsare:θs=6sθm=6(m+s60)θh=30(h+m60+s3600)anglebetweenhourhandandminutehand:Δθhm=30(h+m60+s3600)6(m+s60)Δθhm=30h11m211s120Δθhm=30x3600112(x6060x3600)11120(x60x60)Δθhm=360x360011x120suchthatthisangleis120°,360×x360011x120=±120,±240sayx=3600h+kwith0h11,0k<3600360h11(3600h+k)120=±120,±240k=12011(30h120,240)therearefollowingsolutions:hΔθhm+120120+2402400:21:5401143:420111:27:1801149:60112:32:4801154:360113:38:120114:0:043:420115:5:3001149:60116:10:6001154:360117:16:240118:21:540110:09:27:180115:3001110:32:4801110:6001111:38:1201116:24011thatmeansthereare22instantsatwhichtheanglebetweenhourhandandminutehandis120°.anglebetweenminutehandandsecondhand:Δθms=6(m+s60)6sΔθms=6m59s10Δθms=6(x6060x3600)5910(x60x60)Δθms=360x3600+360x6059x10360x3600+360x6059x10=±120,±240sayx=3600h+60m+kwith0h11,0m59,0k<60360h+360(60h+m)59(3600h+60m+k)10=±120,±2406m59k10=±120,±240for6m59k10=120:60m1200=59k0m2060m1200=59k<59×60m<7920m59k=1059(6m120)for6m59k10=120:60m+1200=59k0m2060m+1200=59k<59×60m<390m38k=1059(6m+120)for6m59k10=240:60m2400=59k0m4060m2400=59k<59×60m<9940m59k=1059(6m240)for6m59k10=240:60m+2400=59k0m4060m+2400=59k<59×60m<190m18k=1059(6m+240)thatmeanstherearetotally40+39+20+19=118instantsatwhichtheanglebetweenminutehandandsecondhandis120°.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com