| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Hello,
here is the problem:
"How long after 7 o'clock will the second hand bisect
the angle formed by the other two hands?"
Plase note that we are interested in the bisection of the acute
angle.
Enjoy,
Kostas G.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 591.1 | Use angles (no trig!) | GALLO::EKLUND | Dave Eklund | Mon Sep 29 1986 11:01 | 20 |
Let A be the angle (in degrees) through which the second hand
travels, B the angle for the Minute hand, and C the angle for the
hour hand. Then B = 1/60 * A and C = 1/3600 * A.
When bisected by the second hand, the two angles are equal, expressed
by (note that 7:00 is 210 degrees, 12:00 is 360 degrees):
A - (210+C) = (360-A) + B
Substituting for B and C from above we get:
A - (210 + 1/3600 * A) = 360 - A + 1/60 * A
Solving for A we get:
A=2052000/7139 degrees or 287.+ degrees.
This translates into 7:00:47.8+ for the time. Incidently, the two
equal angles are about 77 degrees.
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