Question

At what angle are the hands of a clock inclined at $25$ minutes past $5$?

Answer 1

Hint- In order to deal with this issue, we will use the equation as the angle traced by the hour hand is determined by dividing \[{360^0}\] by $12$ hours and multiplying it by the given hour angle traced by the minutes hand by dividing \[{360^0}\] by $60$ minutes and multiplying it by the given minutes.

Complete step-by-step answer:
Given time is $25$ minutes past $5$
As we know that \[{\mathbf{360}}/{\mathbf{12}}{\text{ }} = {\text{ }}{\mathbf{30}}.\]

• The hour hand covers \[{360^0}\] in $12$ hours. So, it’ll cover ${30^0}$ in $1$ hour.

• The minute hand covers \[{360^0}\] in $60$ minutes. So, it’ll cover ${6^0}$ in $1$ min.
Converting given time in hours
$5$ hours $25$minutes \[ = {\text{ }}5{\text{ }} + {\text{ }}25/60{\text{ }} = {\text{ }}5{\text{ }} + {\text{ }}5/12{\text{ }} = {\text{ }}65/12\] hours
As we know that hour hand covers ${30^0}$ in $1$ hour.
So \[65/12 \times {30^0}{\text{ }} = {\text{ }}325/2{\text{ }} = {\text{ }}162.5\] degrees
Similarly for minutes hand
As we know that minutes hand covers ${6^0}$ in $1$ min.
So the angle traced by minutes hand \[ = {\text{ }}25{\text{ }} \times {\text{ }}{6^0}{\text{ }} = {\text{ }}{150^0}\]
The required angle, therefore, between the two hands = angle covered by hour hand-angle covered by minutes hand
\[ = {\text{ }}{162.5^0}{\text{ }} - {\text{ }}{150^0}{\text{ }} = {\text{ }}{12.5^0}\]
Hence the required answer is \[{12.5^0}\]

Note- A clock is a \[{360^0}\] circle, and that each number reflects an angle, with the difference between them being \[{\mathbf{360}}/{\mathbf{12}}{\text{ }} = {\text{ }}{\mathbf{30}}.\]