# 7. The big hand of a clock which measures 26.5 cm long and a small needlemeasures 15 inches long. Determine the distance between the ends of the needles at 2:30.The tenth round of the nearest...

7. The big hand of a clock which measures 26.5 cm long and a small needle

measures 15 inches long. Determine the distance between the ends of the needles at 2:30.

The tenth round of the nearest centimeter

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### 1 Answer

At 2:30, the hour hand (shorter) is between 2 and 3, while the minute hand (longer) is at 6. The angle between numbers is `30^o` . From the midpoint of 2 and 3 to 3, we have `15^o` . Then, from 3 to 6 we have `3*30^o = 90^o` . Hence, the angle between the long and short hand is `90^0 + 15^o = 105^o` .

The length of the two hands in centimeters are `26.5cm` and `15*2.54 = 38.1 cm` . To get the length of the distance between the ends of the two hands, we simply use the Law of Cosine:

`c^2 = a^2 + b^2 - 2abcos(C)` where C is the included angle:

`c^2 = 26.5^2 + 38.1^2 - 2(26.5)(38.1)cos(105)`

`c^2 = 2676.493`

`c = 51.73cm`

Hence, the distance between the two ends is 51.73 cm.