Question

Find the angle between hour-hand and minute hand in a clock at “ten past eleven”.

Answer 1

Hint: Time ten past eleven is ten minutes after ${11}^{\prime }oclock$, hence we have to find the angle between hour-hand and minute-hand when the time in clock is $11.10$ . For finding the angle between two hands of the clock first find the angle covered by both the hands in $1$ minute.

Complete step by step solution:
Firstly we will find the angle covered by both the hands of the clock in $1$ minute.
In $1$ hour or $60$ minutes , minute hand covers distance $={360}^{\circ }$
$\therefore$ In $1$ minute the distance covered $={\displaystyle \frac{{360}^{\circ }}{{60}^{\circ }}}=6^{\circ }$
And In $12$ hours, hour hand cover distance$={360}^{\circ }$
So, in $1$ hour it covers $={\displaystyle \frac{{360}^{\circ }}{12}}={30}^{\circ }$
$\therefore$ in $1$ minute, hour hand covers distance $={\displaystyle \frac{{30}^{\circ }}{{60}^{\circ }}}={\displaystyle \frac{1}{2}}={0.5}^{\circ }$ (as $1hour=60min$ )
Now, we got the angles covered by both the hands of the clock in $1$ minute.
We have to find the angle between both the hands of the clock at $11.10$.
So, at ${11}^{\prime }oclock$ the angle is ${30}^{\circ }$ as hour hand moves ${30}^{\circ }$ in $1$ hour.
So, difference in hour hand and minute hand in $1$ minute$=6^{\circ }-{0.5}^{\circ }={5.5}^{\circ }$
So, in $10$ minutes angle difference $=10\times {5.5}^{\circ }={55}^{\circ }$
Now, total angle between hour and minute hand at $$11.10={30}^{\circ }+{55}^{\circ }={85}^{\circ }$$

Thus the angle between both hands of the clock is ${85}^{\circ }$.

Note: Both the hands of the clock cover ${360}^{\circ }$ in one complete revolution. Hour hand covers ${30}^{\circ }$ in one hour and ${0.5}^{\circ }$ in one minute while the minute hand covers $6^{\circ }$ in one minute.