8 : 2 5 am, what is the measure of the smaller angle between the minute hand and the hour hand?
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1 revolution in an analog clock in 6 0 minutes. One revolution is equivalent to 3 6 0 ° . So the rate of the minute hand is 6 0 m i n 3 6 0 ° = 6 ° p e r m i n u t e . The hour hand completes 1 revolution in an analog clock in 1 2 hours. And 1 2 hours is equivalent to 7 2 0 minutes. So the rate of the hour hand is 7 2 0 m i n 3 6 0 ° = 0 . 5 ° p e r m i n u t e .
The minute hand completesIn short the hour hand moves 0 . 5 ° per minute and the minute hand moves 6 ° per minute.
At 8 : 2 5 the hour hand has move 8 hours and 2 5 minutes taking 0 as the origin.
At 8 : 2 5 the minute hand has move 2 5 minutes taking 0 as the origin.
Therefore,
the hour hand has move ( 8 ) ( 6 0 ) + 2 5 = 5 0 5 minutes or 5 0 5 m i n m i n 0 . 5 ° = 2 5 2 . 5 °
the minute hand has move 2 5 minutes or 2 5 m i n m i n 6 ° = 1 5 0 °
smaller angle formed = 2 5 2 . 5 − 1 5 0 = 1 0 2 . 5 °
If we don't have a handy formula for the angle between hour hand and minute hand, we can see that from digit 5 on the clock to digit 8 it is an angle of 9 0 ∘ .
The hour hand travels h = 1 2 1 × 3 6 0 ∘ = 3 0 ∘ in an hour.
The fraction of the hour is indicated by the minute hand to be f = 6 0 2 5 = 1 2 5 .
In that amount of time the hour hand traveled f × h = 1 2 5 × 3 0 ∘ = 1 2 . 5 ∘ .
At the start of the hour the hour hand was at digit 8, that is 9 0 ∘ past digit 5. Now it is 1 2 . 5 ∘ more.So it is at an angle 9 0 ∘ + 1 2 . 5 ∘ = 1 0 2 . 5 ∘ .
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W e c a n u s e t h e f o r m u l a ∗ ∗ 3 0 h − 2 1 1 m ∗ ∗ f o r f i n d i n g t h e a n g l e b e t w e e n m i n u t e h a n d a n d h o u r h a n d H e r e , h = 8 a n d m = 2 5 ⇒ 3 0 h − 2 1 1 m = 3 0 × 8 − 2 1 1 × 2 5 = 2 4 0 − 1 3 7 . 5 = = 1 0 2 . 5 ∘
∴ A n g l e = 1 0 2 . 5 ∘