Finding Angle
In this article we will study how to find the angle made by the two hands of a clock at a particular time.
So, Time will be given and we will be supposed to find the angle.
To do so we need to have a firm grasp on the Concepts of Angular Distance and Speed.
Concept of Angular Distance & Angular Speed
A full circle = $360^o$
One sector = $30^o$
In case of clocks we often measure the distance in degrees and speed in degrees/minutes.
What is the angular speed of the minute hand?
The minute hand rotates one full circle in one hour, or it travels 360 degrees in 60 minutes.
Therefore, the speed of the minute hand = $360^o$ per 60 minutes = $60^o$ per minute
What is the angular speed of the hour hand?
The hour hand travels $360^o$ in 12 hours.
So, speed of the hour hand = $360^o$ per 12 hours = $30^o$ per 1 hour or per 60 minutes = $0.5^o$ per minute
What is the angular speed of the second hand?
It travels $360^o$ in one minute. So, speed of the second hand = $360^o$ per minute
Q. What is the angle traced by hour hand in 15 minutes?
(a) $7.5^o$ (b) $8^o$ (c) $6.5^o$ (d) $8.5^o$
Explanation:
That is, distance travelled by hour hand in one minute = $0.5^o$
So, angle traced by hour hand in 15 min = 15 × 0.5 = $7.5^o$
Answer: (a)
Finding Angle
In some type of questions, candidate is required to find out the angle between the hour hand and the minute hand at some given particular time.
To solve such questions we can use multiple methods. Let us learn them by taking help of some examples. You are supoosed to choose the method you like and then gain mastery over it.
Q. What is the angle between hour and minute hands at 6:30 pm?
(a) $30^o$ (b) $15^o$ (c) $10^o$ (d) $12^o$
Explanations :
We can visualize that at 6:30 pm the minute hand will be pointing towards 6 in the dial, i.e. where the hour hand was pointing at 6 pm.
Hence, the angle between the two hands will be equal to the distance travelled by the hour hand in 30 minutes, i.e. $0.5^o$ × 30 = $15^o$. (as angular speed of hour hand = $0.5^o$ per minute)
Answer: (b)
The distance between the two hands at 6 pm = $180^o$ (they are facing opposite to each other)
In 30 minutes the minute hand will move by $6^o$ × 30 = $180^o$
In 30 minutes the hour hand will move by $0.5^o$ × 30 = $15^o$
So, the minute hand will now be at the initial position of the hour hand, i.e. it will be pointing towards 6 in the dial. But in this time the hour hand has already moved by $15^o$.
So angle between hour and minute hands at 6:30 pm is $15^o$.
Answer: (b)
We can solve this question using the concept of relative speed. In fact, its’ a faster and shorter method to solve such questions.
The distance between the two hands at 6 pm = $180^o$. So, the minute hand is $180^o$ behind the hour hand.
The relative distance that the minute hand will cover in 30 minutes = $5.5^o$ × 30 = $165^o$.
So angle between hour and minute hands at 6:30 pm = $180^o$ - $165^o$ = $15^o$.
Answer: (b)
We may also make use of a formula. Though the use of formulas should be discouraged as they are applicable only in a few specific settings only.
If the time is A hours and B minutes, then the angle between the two hands of the clock = |(A × 30) – $\frac{(B × 11)}{2}$|
(it’s modulus, i.e. we will always take positive value)
So, at 6:30 pm, the angle between the two hands of the clock = |(6 × 30) – $\frac{(30 × 11)}{2}$| = 180 – 165 = $15^o$.
Answer: (b)