Problem of the week #20 - August 13th, 2012

[Jameson]

Given that the time is 1:52 on an analog clock, calculate the angle between the hour and minute hands (the smaller one).

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[Jameson]

Congratulations to the following members for their correct solutions:

1) Sudharaka
2) soroban
3) BAdhi
4) veronica1999 (no work shown, but I'll give her the benefit of the doubt this time )

Solution (from soroban):

Spoiler

Note from Jameson: There are twelve hour marks on the clock. \(\displaystyle \frac{360^{\circ}}{12}=30 ^{\circ}\) means between each consecutive hour mark there is 30 degrees (between 12-1, 1-2, etc.) This is where the $30^{\circ}$ comes from in his final calculation.


Let [tex]M[/tex] = minute hand, [tex]H[/tex] = hour hand.

At exactly 1:00, [tex]M[/tex] is on "12"; [tex]H[/tex] is on "1".

By 1:52, [tex]M[/tex] has moved [tex]\tfrac{52}{60} = \tfrac{13}{15}[/tex] of the way around the clock.

Then [tex]M[/tex] has moved [tex]\tfrac{13}{15} \times 360^o \,=\,312^o[/tex]

. . Hence, [tex]M[/tex] is [tex]48^o[/tex] from "12".


Meanwhile, [tex]H[/tex] has moved [tex]\tfrac{13}{15}[/tex] of the distance between "1" and "2".
Hence, [tex]H[/tex] is [tex]\tfrac{13}{15}\times 30^o \,=\,26^o[/tex] from "1".


The angle between the hands is: .[tex]48^o + 30^o + 26^o \:=\:104^o.[/tex]