Math Clock Problem?

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 <–––––––

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

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

An old-fashioned clock 
An analog clock 
90° at 3:00 
90° at 9:00

answer is wind it up so it works correctly..

12 hours after that

9:00 is the next time they form a right angle.

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...

The ext time would be 4:05.

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.