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'.

 Measurement of Angles





11 and 12 Grade Math

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