Length of an Arc

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.

3. If α₁, α₂, α₃ radians be the angles subtended by the arcs of lengths l₁, l₂, l₃ at the centers of the circles whose radii are r₁, r₂, r₃ respectively then show that the angle subtended at the centre by the arc of length (l₁ + l₂ + l₃) of a circle whose radius is (r₁ + r₂ + r₃) will be (r₁ α₁ + r₂α₂ + r₃α₃)/(r₁ + r₂ + r₃) radian.

Solution:

According to the problem, the length of an arc l₁ of a circle of radius r₁ subtends an angle α₁ at its center. Hence, using the formula, s = rθ we get,

l₁ = r₁α₁.

Similarly, l₂ = r₂α₂

and l₃ = r₃ α₃.

Therefore, , l₁ + l₂ + l₃ = r₁α₁ + r₂α₂ + r₃α₃.

Let an arc of length (l₁ + l₂ + l₃) of a circle of radius (r₁ + r₂ + r₃) subtend an angle α radian at its center.

Then, α = (l₁ + l₂ + l₃)/(r₁ + r₂ + r₃)

Now, put the value of l₁ = r₁α₁, l₂ = r₂α₂ and l₃ = r₃α₃.

or, α = (r₁α₁ + r₂α₂ + r₃α₃)/(r₁ + r₂ + r₃) radian. Proved.

To solve more problems on length of an arc follow the proof on 'Theta equals s over r'.

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● Measurement of Angles

• Sign of Angles
  
• Trigonometric Angles
• Measure of Angles in Trigonometry
• Systems of Measuring Angles
• Important Properties on Circle
• S is Equal to R Theta
• Sexagesimal, Centesimal and Circular Systems
• Convert the Systems of Measuring Angles
• Convert Circular Measure
• Convert into Radian
• Problems Based on Systems of Measuring Angles
• Length of an Arc
• Problems based on S R Theta Formula

11 and 12 Grade Math

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