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

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#### Jameson

Staff member
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

Staff member
Congratulations to the following members for their correct solutions:

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

Solution (from soroban):

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 $$M$$ = minute hand, $$H$$ = hour hand.

At exactly 1:00, $$M$$ is on "12"; $$H$$ is on "1".

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

Then $$M$$ has moved $$\tfrac{13}{15} \times 360^o \,=\,312^o$$

. . Hence, $$M$$ is $$48^o$$ from "12".

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

The angle between the hands is: .$$48^o + 30^o + 26^o \:=\:104^o.$$

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