Shortly after 10 'o clock, the hands of a clock made a 90 degree angle. What's the time?
Let M be the number of minutes after 10
The hour hand moves (1/2)° every minute
The minute hand moves 6° every minute
Call 10 o'clock = 0°
So....the minute hand is at 60° at 10
So.....we just need to solve this
6M - (1/2)M + 60 = 90
5.5M = 30
M = 30 / 5.5 ≈ 5.45 minutes after 10 = 5 + .45 *60 ≈ 5 min 27 sec after 10
That doesn't make sense! If the minute moves 30 degrees per 5 minute interval, then 30*10 = 300 degrees should be the degrees the minute hand travels until 10!
Shortly after 10 'o clock, the hands of a clock made a 90-degree angle. What's the time?
Here's another way to solve this problem.
Arc from 10 to 12 = 60º
Let x be the arc-path of minute hand in degrees.
60 + x - (x/12) = 90
x = 360/11 = 32.72727272º (minute hand arc)
x/12 = 30/11 = 2.72727272º (hour hand arc)
This is amazing. I just posted an answer to this clock problem on a more recent posting of the problem. I had not seen Phil's solution, but I could have believably copied his verbatim. The same thing happened on a regular decagon problem, even though our solutions were not quite so similar. This is fun; post problems as many times as your heart desires, even though I am getting dizzy looking through the solutions. Just kidding. Please have mercy. I really think only one posting should be allowed at one time, so no time is wasted and the better solutions can be compared.By the way, I tried to sign up today, but the activation email never arrived. Still waiting patiently.