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):
[sp] Note from Jameson: There are twelve hour marks on the clock. [math]\frac{360^{\circ}}{12}=30 ^{\circ}[/math] 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]
. . [/color]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: .[/color][tex]48^o + 30^o + 26^o \:=\:104^o.[/tex]
[/size] [/sp]