1. ## clock problem

At 2 o'clock the angle between hour hand and minute hand is 60
degrees. When will be the angle between hour hand and minute hand 90
degrees?

hour hand moves 0.5 degrees in 1 second and minute hand moves 6
degrees in 1 second.There are 12 digits and eachof them is 30 degrees.
In 1 minute minute hand moves 6 degrees.Hour hand is 1/12 of minute
hand and it is 0.5.
The angle between hour hand and minute hand increases 5.5 degrees in 1
minute. This is how I got it:6-0.5 is 5.5 degrees.

2. ## Re: clock problem

Originally Posted by kastamonu
At 2 o'clock the angle between hour hand and minute hand is 60
degrees. When will be the angle between hour hand and minute hand 90 degrees?
Do not make it so hard.It really is quite simple.
There are $360^o$ around the clock face. At 2 o'clock the hour hand has moved one-sixth of that amount or $60^o$.
So when will the central angle be be $90^o~?$

3. ## Re: clock problem

(1/10 - 1/120) deg/sec ? sec = 30 deg

4. ## Re: clock problem

Originally Posted by kastamonu
At 2 o'clock the angle between hour hand and minute hand is 60
degrees. When will be the angle between hour hand and minute hand 90
degrees?

hour hand moves 0.5 degrees in 1 second and minute hand moves 6
degrees in 1 second.There are 12 digits and eachof them is 30 degrees.
In 1 minute minute hand moves 6 degrees.Hour hand is 1/12 of minute
hand and it is 0.5.
The angle between hour hand and minute hand increases 5.5 degrees in 1
minute. This is how I got it:6-0.5 is 5.5 degrees.
The answer expected is in terms of time. An intuitively obvious answer is 3 o'clock. Certainly the angle is 90 degrees at that time. However, this is not a proof that it is the earliest time.

I should point out that the question (as posed in English) is ambiguous. Assume for now that the angle is to be measured counter-clockwise from the hour hand to the minute hand.

The rate at which the minute hand is moving is $\dfrac{360^o}{hour} = \dfrac{6^o}{minute}.$

The rate at which the hour hand is moving is $\dfrac{360^o}{24\ hours} = \dfrac{0.5^0}{minute}.$

The rate at which the angle is changing is, as you correctly determined, $\dfrac{5.5^o}{min}.$

So it takes $\dfrac{60}{5.5}$ minutes to go from 60 degrees to 0 degrees.

It then takes $\dfrac{360 - 90}{5.5} = \dfrac{270}{5.5}$ minutes to go from 360 degrees to 90 degrees.

Total time = $\dfrac{60}{5.5} + \dfrac{270}{5.5} = \dfrac{330}{5.5} = 60$ minutes, taking us to 3:00 o'clock EXACTLY.

Unfortunately, that is not the only possible way to interpret the question. What is the alternative answer?

5. ## Re: clock problem

Hello, kastamonu!

At 2 o'clock the angle between hour hand and minute hand is 60 degrees.
When will the angle between hour hand and minute hand be 90 degrees?

At exactly 2 o'clock, the hands look like this:

Code:
              o o o
o     *     o
o       *       o
o        * 60o    o
*    *
o         *  *      o
o         o         o
o                   o

o                 o
o               o
o           o
o o o

Several minutes later, the hands look like this:

Code:
              o o o
o          o
o               o
o                 o
*
o           *       o
o         o  90o    o
o           *       o
*
o              *  o
o               o
o           o
o o o
The minute hand starts at $0^o$ and moves 6 degrees per minute.
The position of the minute hand is: . $M \:=\:6t$

The hour hand starts at $60^o$ and moves $\tfrac{1}{2}$ degree per minute.
The position of the hour hand is: . $H \:=\:60 + \tfrac{1}{2}t$

When is the minute hand 90 degrees ahead of the hour hand (first time)?

We have: . $6t \;=\;(60 +\tfrac{1}{2}t) + 90$

Solve for $t\!:\;\;t \,=\,\frac{300}{11}$

Answer: . $27\tfrac{3}{11}$ minutes after 2 o'clock.

6. ## Re: clock problem

an alternative way: 60+90=150 and 150.2/11=300/11.

Many thanks JeffM
Many thanks Soroban

7. ## Re: clock problem

You're welcome