Clocks and Calendar - Angle Problem


 
 
Concept Explanation
 

Clocks and Calendar - Angle Problem

Clock: A Clock is a very common device used to know the time at a particular instance. It is generally circular in shape with three hands indicating hours, minutes and second  viz. an hour hand, minute and second hand. The study of the clock is known as horology

Structure of a Clock:  A basic structure of a clock consists of three hands indicating Hours, minutes and seconds. In the figure above the red hand indicates the second, smaller black hand indicates hours and the larger black hand indicates minutes.The total angle around a point is 360 degree so each hand has to traverse an angle of 360 degree to complete one rotation. These 360 degrees are divided into 12 equal divisions. The angle between the consecutive divisions is obtained by dividing the total angle of clock 360 degree by the number of divisions i.e. 12. Each division indicates the hours at that time.

The angle between any two consecutive divisions

frac{360^0}{12}= 30^0

The angular space between any two consecutive divisions for the hours is further divided into five more divisions. These divisions indicate the minutes.The angular distance traversed by the minutes hand in one minute can be calculated by, dividing the 30 degree by five and is known as the angular value of a minute.

Angular value of a minute

frac{30^0}{5}= 6^0

Angle Problems in Clocks: This concept revolves around the calculation of the angular distance traversed by the different hand of the clock. The different hands of the clock move with different speed.The hour hand takes 12 hours to traverse an angle of 360 degree.

Speed of hour hand = 360^{0} in 12 hours

                              = 30^{0} in 1 hour ( 60 min )

                             = left ( frac{1}{2} right )^0; per;minute

Speed of minute hand = 360^{0} in 1 hour ( 60 min )

                                    = 6^{0} per minute

As they are moving in the same direction

So, the relative speed of minute hand over hour hand will be difference of their speed

 =6^{0}-frac{1^{0}}{2}=5frac{1^{0}}{2}; per; minute

This means that the minute hand approaches and races away from the hour hand at a speed of left ( 5frac{1}{2} right )^o per minute.

Ex.1  What will be the angle between the hour hand and the minute hand at 3 : 40 pm?

Sol.  At 3 pm, the angel is 90^{0}.

     In 40 minutes, the minute hand would cover = 40times 5frac{1}{2}=40timesfrac{11}{2}=220^{0} over the hour hand.

So, the new angle = 220^{0}-90^{0}=130^{0}.

Alternate Solution

Ignoring the movement of hour hand and considering hour hand to be fixed at 3, the angle between the hour hand and the minute hand at 3:40 pm will be 150^{0}.

This is considering the fact that every hour mark is at an angle of 30^{0}. from the previous one .

Now, let us account for the movement of hour hand from 3 pm to 3:40 pm.

The time elapsed = 40 minutes.

So, the angle covered by hour hand =frac{1}{2}times40  minutes

                                                          =20^{0}.

Now, it is clear that the movement of hour hand decreases the angle, which otherwise was calculated to be 150^{0}.

So, the correct angle =150^{0}-20^{0}

                                  =130^{0}.

Sample Questions
(More Questions for each concept available in Login)
Question : 1

What will be the angle between the hour hand and the minute hand at 2 : 00 pm.

Right Option : B
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Explanation
Question : 2

What will be the angle between the hour hand and the minute hand at 9 : 00 pm ?

Right Option : A
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Explanation
Question : 3

What will be the angle between the hour hand and the minute hand at 8 : 00 am ?

Right Option : D
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Explanation
 
 
 


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