Difference between revisions of "1955 AHSME Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
− | Using the formula <math>\frac{|(60h-11m)|}{2}</math>, and plugging in 12 for h and 25 for m, we get the angle between to be 222. | + | Using the formula <math>\frac{|(60h-11m)|}{2}</math>, and plugging in 12 for h and 25 for m, we get the angle between to be 222.5<math>^{\circ}</math>. Since the problem wants the smaller angle, we do <math>360-222.5 = 137.5 \Rightarrow 137^{\circ}30'\Rightarrow \fbox{B}</math> |
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==See Also== | ==See Also== | ||
Latest revision as of 12:37, 4 May 2020
Contents
Problem
The smaller angle between the hands of a clock at p.m. is:
Solution
At , the minute hand is at , or . The hour hand moves 'minutes' every hour, or every hour. At every hour, the hour hand moves minutes on the clock every minute. At , the hour hand is at . Therefore, the angle between the hands is , , or .
Solution 2
Using the formula , and plugging in 12 for h and 25 for m, we get the angle between to be 222.5. Since the problem wants the smaller angle, we do
See Also
1955 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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