Question

At what angle are the hands of a clock inclined at $30$ minutes past $6?$
A). $7{\dfrac{1}{2}^ \circ }$
B). $11{\dfrac{1}{2}^ \circ }$
C). ${15^ \circ }$
D). ${25^ \circ }$

Answer 1

Hint: To solve this question we will use the equation as the angle traced by the hour hand is determined by dividing ${360^ \circ }$ by $12$ hours and then we multiply it with the given hour angle. Similarly we will find the part covered by a minute hand in one hour, then multiply it with the given minute angle. Then we will calculate the required angle between two hands.

Complete step-by-step solution:
We know that an hour hand covers ${360^ \circ }$ in $12$ hour. We will now apply the unitary method.
So in hour it will cover
$\dfrac{{360}}{{12}} = {30^ \circ }$
Similarly the minute hand covers ${360^ \circ }$ in
 $60$ minutes.
 So in one minute it will cover
$\dfrac{{360}}{{60}} = {6^ \circ }$ .
Now we will convert the given time in hours. We know that we can convert minutes into hours by dividing it with $60$ .
So we can convert $6$ hours $30$minutes as follow:
$6 + \dfrac{{30}}{{60}}$ .
We will simplify the value:
 $6 + \dfrac{1}{2} = \dfrac{{12 + 1}}{2}$
It gives us value
 $\dfrac{{13}}{2}$ .
As we know that the hour hand covers $30^\circ $ in one hour.
So
$\dfrac{{13}}{2} \times 30$ .
It gives us value
$13 \times 15 = {195^ \circ }$
Similarly for the minute hand, we know that the minute hand covers $6^\circ $ in one minute.
So we have
 $30 \times 6 = {180^ \circ }$ .
So the required angle between hour hand and minute hand is
 ${195^ \circ } - {180^ \circ }$
It gives us value
${15^ \circ }$ .
Hence the correct option is (c) ${15^ \circ }$ .

Note: We should note that there is also an alternate method to solve this question. We know the formula that angle between the hands of a clock i.e.
$\theta = \dfrac{1}{2}\left( {60 \times h - 11 \times m} \right)$, where
 $h$ is the hour hand and
$m$refers to the minute hand.
Here we have
 $h = 6,m = 30$
By putting the values in formula we have:
$\dfrac{1}{2}\left( {60 \times 6 - 11 \times 30} \right)$
On further simplifying we have
$\dfrac{1}{2}\left( {360 - 330} \right)$
It gives us
$\dfrac{{30}}{2} = 15^\circ $.