The examples will help us to understand to how find the length of an arc using the formula of ‘s is equal to r theta’.
Worked-out problems on length of an arc:
1. In a circle of radius 6 cm, an arc of certain length subtends 20° 17’ at the center. Find in sexagesimal unit the angle subtended by the same arc at the center of a circle of radius 8 cm.
Solution:
Let an arc of length be m cm subtends 20° 17’ at the center of a circle of radius 6 cm and α° at the center of a circle of radius 8 cm.
Now, 20° 17’ = {20 (17/60)}°
= (1217/60)°
= 1217π/(60 × 180) radian [since, 180° = π radian]
And α° = πα/180 radian
We know, the formula, s = rθ then we get,
When the circle of radius is 6 cm; m = 6 × [(1217π)/(60 × 180)] ………… (i)
And when the circle of radius 8 cm; m = 8 × (πα)/180 …………… (ii)
Therefore, from (i) and (ii) we get;
8 × (πα)/180 = 6 × [(1217π)/(60 × 180)]
or, α = [(6/8) × (1217/60)]°
or, α = (3/4) × 20° 17’ [since, (1217/60)° = 20° 17’]
or, α = 3 × 5°4’ 15”
or, α = 15° 12’ 45”.
Therefore, the required angle in sexagesimal unit = 15° 12’ 45”.
2. Aaron is running along a circular track at the rate of 10 mile per hour traverses in 36 seconds an arc which subtends 56° at the center. Find the diameter of the circle.
Solution:
One hour = 3600 seconds
One mile = 5280 feet
Therefore, 10 miles = (5280 × 10) feet = 52800 feet
In 3600 seconds Aaron goes 52800 feet
In 1 second Aaron goes 52800/3600 feet = 44/3 feet
Therefore, in 36 seconds the Aaron goes (44/3) × 36 feet = 528 feet.
Clearly, an arc of length 528 feet subtends 56° = 56 × π/180 radian at the center of the circular track. If ‘y’ feet is the radius of the circular track then using the formula s = rθ we get,
y = s/θ
y = 528/[56 × (π/180)]
y = (528 × 180 × 7)/(56 × 22) feet
y = 540 feet
y = (540/3) yards [since, we know that 3 foot = 1 yard]
y = 180 yards
Therefore, the required diameter = 2 × 180 yards = 360 yards.
To solve more problems on length of an arc follow the proof on 'Theta equals s over r'.
● Measurement of Angles
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