In the previous section we saw the cases when AB is a diameter of the circle. In this section we will see those cases when AB is not a diameter.
Consider fig.27.7 below:
1. The first fig(a) is based on what we have learned so far. Let us write it's details:
• AB is a diameter of the circle. P is any point on the circle above AB. Then, based on theorem 27.1, ∠APB will be equal to 90o
• AB is a diameter of the circle. Q is any point on the circle below AB. Then, based on theorem 27.1, ∠AQB will be equal to 90o
2. Next we want to know the results when AB is not a diameter.
• That is., A and B are two points on the circle. But AB does not pass through the centre O.
• So AB is just a chord. This is shown in fig.27.7(b) above.
• As before, P and Q are any points on the circle. We want to know the angles at P and Q. Let us try:
3. Let us first consider point P alone. So, for now, we will consider ΔAPB only. It is shown in fig.27.7(c).
• We must add some more details to this triangle. We will draw it in a new fig. Consider fig.27.8 below:
4. The first fig(a) shows the same ΔAPB in the previous fig.27.7(c). The vertices A, B and P are now connected to the centre O of the circle.
• Consider ΔAOP:
• OA = OP (∵ radii of the same circle)
• So ΔAOP is isosceles. Then base angles will be equal
• Let the base angles be equal to xo
• Then ∠PAO = ∠APO = xo. These are shown in cyan colour
• ∠AOP = (180 -2x)o (∵ sum of interior angles of a triangle is 180o) This angle is shown in magenta colour
5. Consider ΔBOP:
• OB = OP (∵ radii of the same circle)
• So ΔBOP is isosceles. Then base angles will be equal
• Let the base angles be equal to yo
• Then ∠PBO = ∠BPO = yo. These are shown in white colour
• ∠BOP = (180 -2y)o (∵ sum of interior angles of a triangle is 180o) This angle is shown in blue colour
6. Now consider the angles around the centre point O. There are three angles:
• ∠ AOP = (180 -2x)o
• ∠ BOP = (180 -2y)o
• ∠ AOB = co. This is the angle 'subtended by the chord AB' at the centre O
7. We know that, sum of these three angles must be equal to 360o. So we can write:
(180 -2x) + (180 -2y) + c = 360
⇒ 360 -2x -2y +c = 360 ⇒ c = 2x + 2y ⇒ c = 2(x + y) ⇒ (x + y) = c⁄2
8. But (x + y) is our required ∠APB. We find that, it is equal to c⁄2. This result is shown in fig.27.8(b)
It will be more convenient if we can write this result in terms of arcs.
First we will classify the arcs:
• We know that the chord AB cuts the circle into two arcs. To distinguish between the two arcs, we must give them distinct names.
• For that, we mark any two convenient points X and Y as shown in the animation in fig.27.8(c)
• Now, the larger AXB is the major arc and smaller AYB is the minor arc.
• If the chord is a diameter, the two arcs will be equal. Then there will not be any major or minor
■ Whenever we see an arc, we must immediately understand that, there is a hidden arc also present. When the hidden arc is also drawn, we will get a full circle.
■ So the hidden arc is the alternate arc of the one which is already drawn.
• Example: AYB is the alternate arc of AXB
■ The reverse is also true:
• The arc which is already drawn is the 'alternate arc' of the hidden arc
Example: AXB is the alternate arc of AYB
• Another example:
Consider the animation in fig.27.9 below:
• MXN is the alternate arc of MYN
• MYN is the alternate arc of MXN
• When the two alternate arcs are taken together, we get a full circle
So now we know how to distinguish between arcs. We will get back to the main discussion:
• We have to prove that ∠APB is c⁄2 in this case also.
• Fig(c) shows the angles required for the calculations. But they are over crowded. So we will enlarge the required portion. This is shown in fig.27.11 below:
13. The angles are marked in the same pattern that we saw earlier in fig.27.8(a). See steps (4) and (5). Let us write them:
• Consider ΔAOP:
• OA = OP (∵ radii of the same circle)
• So ΔAOP is isosceles. Then base angles will be equal
• Let the base angles be equal to xo
• Then ∠PAO = ∠APO = xo. These are shown in cyan colour
• ∠AOP = (180 -2x)o (∵ sum of interior angles of a triangle is 180o) This angle is shown in magenta colour
14. Consider BOP:
• OB = OP (∵ radii of the same circle)
• So ΔBOP is isosceles. Then base angles will be equal
• Let the base angles be equal to yo
• Then ∠PBO = ∠BPO = yo. These are shown in white colour
• ∠ BOP = (180 -2y)o (∵ sum of interior angles of a triangle is 180o) This angle is shown in blue colour
15. We want the ∠APB.
(i) From the fig.27.10, it is clear that ∠APB = (y-x)
(ii) Now we want to connect it to the central angle of arc AYB
(iii) The central angle of arc AYB is ∠AOB.
(iv) From the fig.27.11, it is clear that ∠AOB = (180-2x) – (180-2y) = 180 -2x -180 +2y = 2y -2x = 2(y-x)
(v) So ∠AOB = 2(y-x) = 2 × ∠APB [∵ from (i), ∠APB = (y-x)]
(vi) But ∠AOB = co, the central angle
(vii) So we can write: c = 2 × ∠APB ⇒ ∠APB = c⁄2.
16. The above result 15(vii) allows us to write the following steps for fig.27.10(b)
(i) We are given an arc AYB. It has a central angle of co
(ii) Draw the alternate arc AXB. Mark a point P any where on the alternate arc AXB
(iii) The original arc AYB will subtend an ∠APB at P
(iv) Where ever be the point P on the arc AXB, the ∠APB will be equal to c⁄2.
17. The four steps above are the same four steps that we wrote in (10). So we proved that those three steps are valid for any 'not so convenient' points also.
Let us see an example:
In the fig.27.12(a) above:
(i) We are given an arc AYB. It has a central angle of 120o
(ii) Draw the alternate arc AXB. Mark a point P any where on AXB
(iii) The original arc AYB will subtend an ∠APB at P
(iv) Where ever be the point P on the arc AXB, the ∠APB will be equal to 120⁄2 = 60o.
■ Figs. (b) and (c) shows that P can be any where on the arc AXB. The angle 60o will not change.
18. The above fact can be used to calculate the angle at P. Let us see some examples:
In fig.27.13(a) above:
(i) We are given an arc AYB. It has a central angle of 86o
(ii) A point P is marked on the alternate arc.
• What is the angle subtended at P, by the original arc AYB?
Solution:
The required angle will be equal to 86⁄2 = 43o
In fig.27.13(b) above:
(i) We are given an arc AYB. It has a central angle of 218o
(ii) A point P is marked on the alternate arc.
• What is the angle subtended at P, by the original arc AYB?
Solution:
The required angle will be equal to 218⁄2 = 109o
In fig.27.13(c) above:
(i) We are given an arc AYB. It has a central angle of 284o
(ii) A point P is marked on the alternate arc.
• What is the angle subtended at P, by the original arc AYB?
Solution:
The required angle will be equal to 284⁄2 = 142o
Consider fig.27.7 below:
 |
| Fig.27.7 |
• AB is a diameter of the circle. P is any point on the circle above AB. Then, based on theorem 27.1, ∠APB will be equal to 90o
• AB is a diameter of the circle. Q is any point on the circle below AB. Then, based on theorem 27.1, ∠AQB will be equal to 90o
2. Next we want to know the results when AB is not a diameter.
• That is., A and B are two points on the circle. But AB does not pass through the centre O.
• So AB is just a chord. This is shown in fig.27.7(b) above.
• As before, P and Q are any points on the circle. We want to know the angles at P and Q. Let us try:
3. Let us first consider point P alone. So, for now, we will consider ΔAPB only. It is shown in fig.27.7(c).
• We must add some more details to this triangle. We will draw it in a new fig. Consider fig.27.8 below:
 |
| Fig.27.8 |
• Consider ΔAOP:
• OA = OP (∵ radii of the same circle)
• So ΔAOP is isosceles. Then base angles will be equal
• Let the base angles be equal to xo
• Then ∠PAO = ∠APO = xo. These are shown in cyan colour
• ∠AOP = (180 -2x)o (∵ sum of interior angles of a triangle is 180o) This angle is shown in magenta colour
5. Consider ΔBOP:
• OB = OP (∵ radii of the same circle)
• So ΔBOP is isosceles. Then base angles will be equal
• Let the base angles be equal to yo
• Then ∠PBO = ∠BPO = yo. These are shown in white colour
• ∠BOP = (180 -2y)o (∵ sum of interior angles of a triangle is 180o) This angle is shown in blue colour
6. Now consider the angles around the centre point O. There are three angles:
• ∠ AOP = (180 -2x)o
• ∠ BOP = (180 -2y)o
• ∠ AOB = co. This is the angle 'subtended by the chord AB' at the centre O
7. We know that, sum of these three angles must be equal to 360o. So we can write:
(180 -2x) + (180 -2y) + c = 360
⇒ 360 -2x -2y +c = 360 ⇒ c = 2x + 2y ⇒ c = 2(x + y) ⇒ (x + y) = c⁄2
8. But (x + y) is our required ∠APB. We find that, it is equal to c⁄2. This result is shown in fig.27.8(b)
First we will classify the arcs:
• We know that the chord AB cuts the circle into two arcs. To distinguish between the two arcs, we must give them distinct names.
• For that, we mark any two convenient points X and Y as shown in the animation in fig.27.8(c)
• Now, the larger AXB is the major arc and smaller AYB is the minor arc.
• If the chord is a diameter, the two arcs will be equal. Then there will not be any major or minor
■ Whenever we see an arc, we must immediately understand that, there is a hidden arc also present. When the hidden arc is also drawn, we will get a full circle.
■ So the hidden arc is the alternate arc of the one which is already drawn.
• Example: AYB is the alternate arc of AXB
■ The reverse is also true:
• The arc which is already drawn is the 'alternate arc' of the hidden arc
Example: AXB is the alternate arc of AYB
• Another example:
Consider the animation in fig.27.9 below:
 |
| Fig.27.9 |
• MYN is the alternate arc of MXN
• When the two alternate arcs are taken together, we get a full circle
9. In the fig.27.8(c), there are no chords. There are only arcs.
• P is a point on the arc AXB.
10. So we can write the result that we obtained in (8) as follows:
(i) We are given an arc AYB. It has a central angle of co
(ii) Draw the alternate arc AXB. Mark a point P any where on the alternate arc AXB
(iii) The original arc AYB will subtend an ∠APB at P
(iv) Where ever be the point P on the arc AXB, the ∠APB will be equal to c⁄2.
• Fig(a) shows the same 'convenient point' P that we used so far.
• In fig(b), P is shifted to a new 'not so convenient' position closer to B. (i) We are given an arc AYB. It has a central angle of co
(ii) Draw the alternate arc AXB. Mark a point P any where on the alternate arc AXB
(iii) The original arc AYB will subtend an ∠APB at P
(iv) Where ever be the point P on the arc AXB, the ∠APB will be equal to c⁄2.
11. We arrived at the above results in (10), through the calculations based on fig.27.8(a).
• However, in fig.27.8(a), we chose a 'convenient' point P. Let us choose a 'not so convenient' point and see whether we can obtain the same results:
12. Consider fig.27.10 below:• However, in fig.27.8(a), we chose a 'convenient' point P. Let us choose a 'not so convenient' point and see whether we can obtain the same results:
 |
| Fig.27.10 |
• We have to prove that ∠APB is c⁄2 in this case also.
• Fig(c) shows the angles required for the calculations. But they are over crowded. So we will enlarge the required portion. This is shown in fig.27.11 below:
 |
| Fig.27.11 |
• Consider ΔAOP:
• OA = OP (∵ radii of the same circle)
• So ΔAOP is isosceles. Then base angles will be equal
• Let the base angles be equal to xo
• Then ∠PAO = ∠APO = xo. These are shown in cyan colour
• ∠AOP = (180 -2x)o (∵ sum of interior angles of a triangle is 180o) This angle is shown in magenta colour
14. Consider BOP:
• OB = OP (∵ radii of the same circle)
• So ΔBOP is isosceles. Then base angles will be equal
• Let the base angles be equal to yo
• Then ∠PBO = ∠BPO = yo. These are shown in white colour
• ∠ BOP = (180 -2y)o (∵ sum of interior angles of a triangle is 180o) This angle is shown in blue colour
15. We want the ∠APB.
(i) From the fig.27.10, it is clear that ∠APB = (y-x)
(ii) Now we want to connect it to the central angle of arc AYB
(iii) The central angle of arc AYB is ∠AOB.
(iv) From the fig.27.11, it is clear that ∠AOB = (180-2x) – (180-2y) = 180 -2x -180 +2y = 2y -2x = 2(y-x)
(v) So ∠AOB = 2(y-x) = 2 × ∠APB [∵ from (i), ∠APB = (y-x)]
(vi) But ∠AOB = co, the central angle
(vii) So we can write: c = 2 × ∠APB ⇒ ∠APB = c⁄2.
16. The above result 15(vii) allows us to write the following steps for fig.27.10(b)
(i) We are given an arc AYB. It has a central angle of co
(ii) Draw the alternate arc AXB. Mark a point P any where on the alternate arc AXB
(iii) The original arc AYB will subtend an ∠APB at P
(iv) Where ever be the point P on the arc AXB, the ∠APB will be equal to c⁄2.
17. The four steps above are the same four steps that we wrote in (10). So we proved that those three steps are valid for any 'not so convenient' points also.
Let us see an example:
 |
| Fig.27.12 |
(i) We are given an arc AYB. It has a central angle of 120o
(ii) Draw the alternate arc AXB. Mark a point P any where on AXB
(iii) The original arc AYB will subtend an ∠APB at P
(iv) Where ever be the point P on the arc AXB, the ∠APB will be equal to 120⁄2 = 60o.
■ Figs. (b) and (c) shows that P can be any where on the arc AXB. The angle 60o will not change.
18. The above fact can be used to calculate the angle at P. Let us see some examples:
 |
| Fig.27.13 |
(i) We are given an arc AYB. It has a central angle of 86o
(ii) A point P is marked on the alternate arc.
• What is the angle subtended at P, by the original arc AYB?
Solution:
The required angle will be equal to 86⁄2 = 43o
In fig.27.13(b) above:
(i) We are given an arc AYB. It has a central angle of 218o
(ii) A point P is marked on the alternate arc.
• What is the angle subtended at P, by the original arc AYB?
Solution:
The required angle will be equal to 218⁄2 = 109o
In fig.27.13(c) above:
(i) We are given an arc AYB. It has a central angle of 284o
(ii) A point P is marked on the alternate arc.
• What is the angle subtended at P, by the original arc AYB?
Solution:
The required angle will be equal to 284⁄2 = 142o
• In the above discussion, we are given the minor arc AYB and a point P is marked on the alternate arc AXB
• What if we are given the major arc AXB and a point P is marked on the alternate arc AYB?
We will see that case in the next section.
• What if we are given the major arc AXB and a point P is marked on the alternate arc AYB?
We will see that case in the next section.
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