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Q.
The angle between two hands of a clock at quarter past one is
Trigonometric Functions
Solution:
If hours hand were at $1$ and minutes hand at $3$, the angle between the two hands would have been $60^{\circ}$. In $15$ minutes, hours hand revolves through $\left( \frac{360 \times 15}{720} \right)^{\circ} = \left(7 \frac{1}{2} \right)^{\circ}$
($\because$ In $12$ hours, i.e., $720$ min, hours hand revolves through $360^{\circ}$)
$\therefore $ Required angle between the hands of clock
$ = 60^{\circ} - \left(7 \frac{1}{2}\right)^{\circ} = \left(52 \frac{1}{2}\right)^{\circ}$
