# Math Clock Problem?

An old fashion clock with twelve numbers on the face and hands forms a right angle at 3:00. What is the next time they form a right angle?

### 10 Answers

- MichaelLv 72 months agoFavourite answer
Hour hand:

makes one revolution in 12 hours

in one minute the angles increases by

360 / (12*60) = 0.5º

It starts at 3:00 which is 90º

ϴ₁ = 90 + 0.5m

----------------------

Minute hand

makes one revoulution in 1 hour

in one minute the angles increases by

360 / 60 = 6º

It starts at 0º

ϴ₂ = 6m

---------------------

The next time they will be at 90º

ϴ₂ - ϴ₁ = 90

6m - (90 + 0.5m) = 90

5.5m = 180

m = 32.7272727273

32 minutes 43.63... seconds

time will be

~ 3:32:44 <–––––––

- Anonymous1 month ago
An old-fashioned clock

An analog clock

90° at 3:00

90° at 9:00

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- micatkieLv 72 months ago
Let y min after 3:00 be the next time when the two hands form a right angle.

The hour hand turns 360°/12 = 30° in 60 min. Hence, in y min it turns 30° × (y/60) = 0.5y°

The minute hand turns 360° in 60 min. Hence, in y min it turns 360° × (y/60) = 6y°

When next time the two hands form a right angle, the minutes hand turns 180° more than hour hand:

6y = 0.5y + 180

5.5y = 180

y = 32 and 8/11

Answer: The time is (32 and 8/11) min ≈ 32.73 min after 3:00. It is about 3:32:44

- UserLv 72 months ago
at about 3:33

But let's calculate exactly (as Captain tried and failed to do).

First, let's remember to put EVERYTHING in the form of minutes.

We know that the hour hand position will be at 3 o'clock + a location equal to 1/60 times 5 minutes (because the hour hand moves the space of 5 minutes in one hour)

H = 3 o'clock + 5 * min/60

where min is the number of minutes past 3 o'clock.

Remember, we're doing everything in minutes, so 3 o'clock is actually 15 min.

(eq 1) H = 15 + min/12

and when the minute hand is at 90 deg to the hour hand it will be 15 minutes ahead of the hour hand position

(eq 2) min = H + 15

Use (eq 1) to replace H in (eq 2)

min = 15 + min/12 + 15

now solve

(min - 30) = min/12

12min - 360 = min

11min = 360

min = 32.7

H = 15 + min/12 = 17.7

which is a position 15 minutes before the minute hand at 32.7 minutes

- 2 months ago
The hour hand moves at 1/60th of the rate of the minute hand. When the minute hand travels t degrees, the hour hand will travel t/60 degrees. The hour hand is at t = 0 when the minute hand is at t = -90

m = -90 + t

h = t/60

We want to know when m = h + 90

t/60 + 90 = -90 + t

180 = t - t/60

180 = 59t/60

180 * 60/59 = t

t = 10800 / 59

t = 183.05084745762711864406779661017....

But what time does this correspond to? Well, one revolution is 360 degrees and it's equal to 60 minutes

60 * t / 360 =>

t/6 =>

30.508474576271186440677966101695

At 3:30:30.50847...., the minute hand and hour hand will be 90 degrees from each other.

It's kind of neat, a sort of recursion has happened. I had to check twice to make sure I hadn't accidentally done something wrong. I had gotten 30.50847.... minutes, so I subtracted 30 from that, which gave me 0.50847.... minutes. I multiplied that by 60 and got 30.50847... seconds. I could keep doing that forever.

- billrussell42Lv 72 months ago
at about 3:31

in more detail:

angle of hour hand with vertical is h(360/12) = 30h (0<h≤12)

where h is hour in hours and decimal minutes

or angle = 30(h+m/60)

angle of minute hand is m(360/60) = 6m (0≤m<60)

you can plug in numbers

90º angle

|30(h+m/60) – 6m| = 90

for h=3

30(3+m/60) – 6m = 90

90 + m/2 – 6m = 90

m/2 – 6m = 0

m = 0 (ie 3 hr 0 min)

or

|30(h+m/60) – 6m| = 90

6m – 30(3+m/60) = 90

6m – 90 – m/60 = 90

6m – m/60 = 180

360m/60 – m/60 = 180

359m/60 = 180

m = 30.1 min

ie 3 hr 30.1 min

but check the math...