What is the measure of the smaller angle formed by the minute and hour hands of a clock at 3:47 ?
less than 180 deg but not exactly 120 degree i need the exact
155 degree? the other hook is not moving haha can u check
\[\angle = \frac{ 47 - (15+\frac{ 47 }{60 } \times 5) }{ 60 } \times 360 \] I think this should give the angle. I'll try to explain this formula...
If you take the simpler case, where the hour hand would be right on the 3, then the formula for the angle becomes:\[\frac{ 47-15 }{ 60 } \times 360\]
A whole hour is 360 degrees, or 60/60 * 360. 30 minutes would be 30/60 * 360, or 180. Here we have the minute hand at 47, and the hour hand is between the 3 and the 4. To find exactly where the hour hand is, start at :15, and add 47/60 * 5 (since there's 5 notches/minutes between :15 and :20 on the clock-face), so the hour hand is right at :18.91666 (or almost 19 minutes past the 12 o'clock). so the angle = (45-18.91666)/60 * 360 degrees.
Turns out there's entire webpages dedicated to clock-angle problems: http://en.wikipedia.org/wiki/Clock_angle_problem https://sites.google.com/site/mymathclassroom/trigonometry/clock-angle-problems/clock-angle-problem-formula The formula they give on these sites gives the same angle as mine does. Mine just looks uglier.
i get 156.5 deg
from your formula
I think you made a mistake, should be... 168.5
(45-18.91666)/60 * 360 degrees. from this?
oh, sorry the 45 there should be 47. It's 3:47 not 3:45. The other formula i gave was correct.
hahaha i didnt notice
168.5 correct 2x
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