1. Help! Angles and Clocks

How would I go about figuring this out?

Mario was studying angles and wondered if during his math class the minute and hour hands of the clock formed a right angle. If his class meets from 6:00 P.M. to 6:50 P.M., is a right angle formed? If so, figure out to the nearest second when the hands form the right angle.

2. Re: Help! Angles and Clocks

What are the initial positions of the two hands, and at what angular speed in degrees per hour do the two hands move?

3. Re: Help! Angles and Clocks

I haven't a clue! Haha. This is all I was given to work with.

4. Re: Help! Angles and Clocks

Using the standard convention for measuring angles on the unit circle, at 6:00 pm, at what angle is the minute hand? The hour hand?

5. Re: Help! Angles and Clocks

Originally Posted by stallioncane
I haven't a clue! Haha. This is all I was given to work with.
FYI: He's giving you hints about how to start.

-Dan

6. Re: Help! Angles and Clocks

Well I've got it down to 6:16 but I can't figure out how to do the seconds.

7. Re: Help! Angles and Clocks

That's close to one of the valid times, but without knowing what you did, I can't really help.

8. Re: Help! Angles and Clocks

Honestly, I'm an early childhood education major. I busted out a Judy Clock and figured it out that way. Is there a particular formula I can use to find the seconds?

9. Re: Help! Angles and Clocks

Nevermind. Figured it out. Thanks!

10. Re: Help! Angles and Clocks

Are you familiar with angles on the unit circle?

11. Re: Help! Angles and Clocks

Hello, stallioncane!

Here is my approach to Clock-hands Problems.

Mario was studying angles and wondered if during his math class
the minute and hour hands of the clock formed a right angle.
If his class meets from 6:00 P.M. to 6:50 P.M., is a right angle formed?
If so, figure out to the nearest second when the hands form a right angle.

You might think that one answer looks like this:
Code:
              * * *
*           *
*               *
*                 *

*                   *
*         *-------- * 3
*         |         *
|
*        |        *
*               *
*           *
* * *
6
But this is incorrect.

By the time the minute hand has moved 90o to the "3",
. . the hour hand has already moved toward the "7".

The minute hand moves 360o in 60 minutes.
It moves $6^o$ per minute.
It starts at "12" (0o) and in the next $t$ minutes, it move $6t$ degrees.
The position of the minute hand at time $t$ is: . $M \:=\:6t$ degrees.

The hour hand moves 30o in 60 minutes.
It move $\tfrac{1}{2}^o$ per minute.
It starts at "6" (180o) and in the next $t$ minutes, it moves $\tfrac{1}{2}t$ degrees.
The position on the hour hand at time $t$ is: . $H \:=\:180 + \tfrac{1}{2}t$ degrees.

There are two possible cases.

[1] The hour hand is 90o ahead of the minute hand: . $H \:=\:M + 90$

. . . $180 + \tfrac{1}{2}t \:=\:6t + 90 \quad\Rightarrow\quad \tfrac{11}{2}t \:=\:90 \quad\Rightarrow\quad t \:=\:\tfrac{180}{11}$

. . . $t \:=\:16\tfrac{4}{11} \:=\:16\text{ minutes, }21.\overline{81}\text{ seconds}$

Time: 16 minutes, 22 seconds after 6 o'clock.

[2] The minute hand is 90o ahead of the hour hand: . $M \:=\:H + 90$

. . . $6t \:=\:180 + \tfrac{1}{2}t 90 \quad\Rightarrow\quad \tfrac{11}{2}t \:=\:270 \quad\Rightarrow\quad t \:=\:\frac{540}{11}$

. . . $t \:=\:49\tfrac{1}{11} \:=\:49\text{ minutes, }5.\overline{45}\text{ seconds}$

Time: 49 minutes, 5 seconds after 6 o'clock.