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- Jan 26, 2012

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Thread starter
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- #1

- Jan 26, 2012

- 4,052

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- #2

- Jan 26, 2012

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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):

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]

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