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 $11'oclock$, 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 $=\dfrac{{{360}^{\circ }}}{{{60}^{\circ }}}={{6}^{\circ }}$
And In $12$ hours, hour hand cover distance$={{360}^{\circ }}$
So, in $1$ hour it covers $=\dfrac{{{360}^{\circ }}}{12}\,\,=\,\,{{30}^{\circ }}$
$\therefore $ in $1$ minute, hour hand covers distance $=\dfrac{{{30}^{\circ }}}{{{60}^{\circ }}}=\dfrac{1}{2}={{0.5}^{\circ }}$ (as $1\,hour\,=\,60\,\min $ )
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 $11'oclock$ 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.