In the previous section we have seen the length of an arc for each 1o turn. In this section we will see the converse. That is., we can find the 'angle turned' for each 1 cm of the arc. We can find this from the same four cases. We will use the same fig.21.30. For convenience, it is shown again below:
Case 1:
■ Two arcs may have the same central angle. But they may have different lengths. Out of the two arcs, the one which gave a greater radius will have a greater length.
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| Fig.21.30 |
1. In fig.21.30(b), when a point travels from P to Q along the minor arc PQ, the distance covered is one fourth of the total perimeter of the ‘circle with radius r cm’.
2. We know that the total perimeter = 2πr cm. So the distance covered = (1⁄4 × 2πr) cm
3. Also, when the above distance is covered, we can say, the point ‘turns’ through an angle of 90o
4. So we can write: (1⁄4 × 2πr) cm → 90o
5. So 1 cm → [90 ÷ (1⁄4 × 2πr)]
6. [90 ÷ (1⁄4 × 2πr)] = (90×4)⁄2πr = 180⁄πr
7. So 1 cm → 180⁄πr. That means, when the point travels a distance of 1 cm along the arc, the angle turned is (180⁄πr)o
8. Another form of writing this is: In a circle of radius r, an arc of length 1 cm will subtend an angle of (180⁄πr)o at the centre.
The two theorems can be represented in diagrams as shown in the fig.21.31 below. Fig.a represents theorem 21.1 and fig.b represents theorem 21.28. Another form of writing this is: In a circle of radius r, an arc of length 1 cm will subtend an angle of (180⁄πr)o at the centre.
Case 2:
1. When a point travels from P to R along the arc PQR, the distance covered is one half of the total perimeter of the ‘circle with radius r cm’.
2. We know that the total perimeter = 2πr. So the distance covered = (1⁄2 × 2πr) cm
3. Also, when the above distance is covered, we can say, the point ‘turns’ through an angle of (90+90) = 180o
4. So we can write: (1⁄2 × 2πr) cm → 180o
5. So 1 cm → [180 ÷ (1⁄2 × 2πr)]
6. [180 ÷ (1⁄2 × 2πr)] = (180×2)⁄2πr = 180⁄πr
7. So 1 cm → 180⁄πr. That means, when the point travels a distance of 1 cm along the arc, the angle turned is (180⁄πr)o
8. Another form of writing this is: In a circle of radius r, an arc of length 1 cm will subtend an angle of (180⁄πr)o at the centre.
Case 3:
7. So 1 cm → 180⁄πr. That means, when the point travels a distance of 1 cm along the arc, the angle turned is (180⁄πr)o
8. Another form of writing this is: In a circle of radius r, an arc of length 1 cm will subtend an angle of (180⁄πr)o at the centre.
Case 3:
1. When a point travels from P to S along the arc PQRS, the distance covered is three fourth of the total perimeter of the ‘circle with radius r cm’.
2. We know that the total perimeter = 2πr. So the distance covered = (3⁄4 × 2πr)
3. Also, when the above distance is covered, we can say, the point ‘turns’ through an angle of (90+90+90) = 270o
4. So we can write: (3⁄4 × 2πr) cm → 270o
5. So 1 cm → [270 ÷ (3⁄4 × 2πr)]
6. [270 ÷ (3⁄4 × 2πr)] = (270×4)⁄6πr = 180⁄πr
7. So 1 cm → 180⁄πr. That means, when the point travels a distance of 1 cm along the arc, the angle turned is (180⁄πr)o
8. Another form of writing this is: In a circle of radius r, an arc of length 1 cm will subtend an angle of (180⁄πr)o at the centre.
Case 4:
7. So 1 cm → 180⁄πr. That means, when the point travels a distance of 1 cm along the arc, the angle turned is (180⁄πr)o
8. Another form of writing this is: In a circle of radius r, an arc of length 1 cm will subtend an angle of (180⁄πr)o at the centre.
Case 4:
1. When a point travels from P, and returns back to P along the arc PQRSP, the distance covered is one full of the total perimeter of the ‘circle with radius r cm’.
2. We know that the total perimeter = 2πr. So the distance covered = (2πr) cm
3. Also, when the above distance is covered, we can say, the point ‘turns’ through an angle of (90+90+90+90) = 360o
4. So we can write: (2πr) cm → 360o
5. So 1 cm → [360 ÷ (2πr)]
6. [360 ÷ (2πr)] == 180⁄πr
7. So 1 cm → 180⁄πr. That means, when the point travels a distance of 1 cm along the arc, the angle turned is (180⁄πr)o
8. Another form of writing this is: In a circle of radius r, an arc of length 1 cm will subtend an angle of (180⁄πr)o at the centre.
■ In all the four cases, we get the same result. We can write it in the form of a theorem:
Theorem 21.2:
• A point is situated on the circumference of a circle with radius r
• It travels a distance of 1 cm along the circumference of the circle
• Then the angle turned by the point is (180⁄πr)o.
• Another form of writing this is: If the radius of a circle is r cm, then an arc of length l cm will subtend an angle of (180⁄πr)o at the centre.
• So, if the length of an arc (in a circle of radius r) is 'l' cm, the angle subtended by that arc at the centre
= (180⁄πr × l)o = (180l⁄πr)o.
Based on the above theorem, we can write:
In fig.21.30(a), if the length of the arc AB is l, then x = (180l⁄πr)o.
A sample calculation:
In a circle of radius 3 cm, length of an arc is 2.5 cm. What is the central angle of the arc?
Solution:
1. According to theorem 21.2 above, In a circle of radius r, for every 1 cm length of an arc, the central angle will be equal to (180⁄πr)o.
2. So in a circle of radius 3 cm, an arc of length 2.5 cm will make a central angle of
[(180⁄3π) × 2.5] = [(60⁄π) × 2.5] = (150⁄π) = 47.77o
7. So 1 cm → 180⁄πr. That means, when the point travels a distance of 1 cm along the arc, the angle turned is (180⁄πr)o
8. Another form of writing this is: In a circle of radius r, an arc of length 1 cm will subtend an angle of (180⁄πr)o at the centre.
■ In all the four cases, we get the same result. We can write it in the form of a theorem:
Theorem 21.2:
• A point is situated on the circumference of a circle with radius r
• It travels a distance of 1 cm along the circumference of the circle
• Then the angle turned by the point is (180⁄πr)o.
• Another form of writing this is: If the radius of a circle is r cm, then an arc of length l cm will subtend an angle of (180⁄πr)o at the centre.
• So, if the length of an arc (in a circle of radius r) is 'l' cm, the angle subtended by that arc at the centre
= (180⁄πr × l)o = (180l⁄πr)o.
Based on the above theorem, we can write:
In fig.21.30(a), if the length of the arc AB is l, then x = (180l⁄πr)o.
A sample calculation:
In a circle of radius 3 cm, length of an arc is 2.5 cm. What is the central angle of the arc?
Solution:
1. According to theorem 21.2 above, In a circle of radius r, for every 1 cm length of an arc, the central angle will be equal to (180⁄πr)o.
2. So in a circle of radius 3 cm, an arc of length 2.5 cm will make a central angle of
[(180⁄3π) × 2.5] = [(60⁄π) × 2.5] = (150⁄π) = 47.77o
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| Fig.21.31 |
Now we will see a very
interesting case:
• In the fig.21.32, two circles with
radii r1 cm and r2 cm are drawn.
• An arc AB is marked in the inner circle.
And an arc PQ is marked in the outer circle.
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| Fig.21.32 |
• Both
arcs have the same central angle xo. Using this central angle, we can calculate their lengths:
1. We have: Length of arc AB (Theorem 21.1) = (πr1x⁄180) cm
2. Length of arc PQ = (πr2x⁄180) cm
3. We find that the lengths are
different. Let us compare them: (πr1x⁄180) and (πr2x⁄180) cm
4. The only difference is in the
radius. r2 is greater than r1. So the length of arc PQ will be
greater than the length of arc AB.
So we can write: ■ Two arcs may have the same central angle. But they may have different lengths. Out of the two arcs, the one which gave a greater radius will have a greater length.
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