In the previous sections we saw theorem 27.4 and it's converse. We also saw some solved examples.
■ So far we have been considering the following case:
• An arc subtends an angle at a point P on the alternate arc.
♦ That means, the point P is situated some where on the circle
■ Now we consider the other case:
• An arc subtends an angle at a point P which is not on the circle.
• Such a case is shown in the fig.27.32(a) below:
• In this case, P is in the interior of the circle. We want to know the ∠APB
For that, the following steps are used:
1. Extend PA along the same line, to meet the circle at M
2. Extend PB along the same line, to meet the circle at N
3. Name arc AB as AXB and arc MN as MYN. This is shown in fig(b)
4. Let the central angle of arc AXB be xo. This is shown in fig(c)
5. So, if this arc AXB subtends an angle at any point on it's alternate arc, that angle would be x⁄2.
Is there any such point?
Indeed there is.
6. If we join A to N, we will see that, arc AXB subtends an ∠ANB at the point N on the alternate arc.
So ∠ANB = x⁄2. This is shown in fig(d) below:
7. Let the central angle of arc MYN be yo
8. So, if this arc MYN subtends an angle at any point on it's alternate arc, that angle would be y⁄2
Is there any such point?
Indeed there is.
9. Arc MYN subtends an ∠MAN at the point A on the alternate arc.
So ∠MAN = y⁄2
10. Now consider ΔPAN in fig(e).
• ∠ APB is an exterior angle in PAN
• Exterior angle = sum of remote interior angles.
11. So we get: ∠APB = x⁄2 + y⁄2 ⇒ APB = (x+y)⁄2
We can write the above result in the form of a theorem. We will write it in steps:
2. Let it subtend an ∠APB at a point P in the interior of the circle
3. Also let the extensions of the two legs AP and BP subtend an arc MYN on the circle
4. Then ∠APB is the average of the following two items:
(i) Central angle of arc AXB
(ii) Central angle of arc MYN
1. Let the leg AP intersect the circle at M
2. Let the leg BP intersect the circle at N
3. Name arc AB as AXB and arc MN as MYN. This is shown in fig(b)
4. Let the central angle of arc AXB be xo. This is shown in fig(c)
5. So, if this arc AXB subtends an angle at any point on it's alternate arc, that angle would be x⁄2
Is there any such point?
Indeed there is.
6. If we join A to N, we will see that, arc AXB subtends an ∠ANB at the point N on the alternate arc.
So ANB = x⁄2. This is shown in fig(d)
7. Let the central angle of arc MYN be y
8. So, if this arc MYN subtends an angle at any point on it's alternate arc, that angle would be y⁄2
Is there any such point?
Indeed there is.
9. Arc MYN subtends an ∠MAN at the point A on the alternate arc.
So ∠MAN = y⁄2
10. Now consider PAN in fig(e).
• ∠ ANB is an exterior angle in ΔPAN
• Exterior angle = sum of remote interior angles.
11. So we get: ∠ANB = ∠APB + ∠PAN ⇒x⁄2 = ∠APB + x⁄2 ⇒ ∠APB = (x-y)⁄2
We can write the above result in the form of a theorem. We will write it in steps:
2. Let it subtend ∠APB at a point P in the exterior of the circle
3. Also let the two legs AP and BP intersect the circle at M and N respectively
4. Then ∠APB is calculated as follows:
(i) Write the central angle of arc AXB
(ii) Write the central angle of arc MYN
(iii) Subtract the smaller from the larger
(iv) Half of the result of subtraction will be ∠APB
Now we will see some solved examples
■ So far we have been considering the following case:
• An arc subtends an angle at a point P on the alternate arc.
♦ That means, the point P is situated some where on the circle
■ Now we consider the other case:
• An arc subtends an angle at a point P which is not on the circle.
• Such a case is shown in the fig.27.32(a) below:
 |
| Fig.27.32 |
For that, the following steps are used:
1. Extend PA along the same line, to meet the circle at M
2. Extend PB along the same line, to meet the circle at N
3. Name arc AB as AXB and arc MN as MYN. This is shown in fig(b)
4. Let the central angle of arc AXB be xo. This is shown in fig(c)
5. So, if this arc AXB subtends an angle at any point on it's alternate arc, that angle would be x⁄2.
Is there any such point?
Indeed there is.
6. If we join A to N, we will see that, arc AXB subtends an ∠ANB at the point N on the alternate arc.
So ∠ANB = x⁄2. This is shown in fig(d) below:
 |
| Fig.27.32 |
8. So, if this arc MYN subtends an angle at any point on it's alternate arc, that angle would be y⁄2
Is there any such point?
Indeed there is.
9. Arc MYN subtends an ∠MAN at the point A on the alternate arc.
So ∠MAN = y⁄2
10. Now consider ΔPAN in fig(e).
• ∠ APB is an exterior angle in PAN
• Exterior angle = sum of remote interior angles.
11. So we get: ∠APB = x⁄2 + y⁄2 ⇒ APB = (x+y)⁄2
Theorem 27.5
1. Consider any arc AXB2. Let it subtend an ∠APB at a point P in the interior of the circle
3. Also let the extensions of the two legs AP and BP subtend an arc MYN on the circle
4. Then ∠APB is the average of the following two items:
(i) Central angle of arc AXB
(ii) Central angle of arc MYN
Another case is shown in the fig.27.33(a) below:
• In this case, P is in the exterior of the circle. We want to know the ∠APB
For that, the following steps are used: |
| Fig.27.33 |
1. Let the leg AP intersect the circle at M
2. Let the leg BP intersect the circle at N
3. Name arc AB as AXB and arc MN as MYN. This is shown in fig(b)
4. Let the central angle of arc AXB be xo. This is shown in fig(c)
5. So, if this arc AXB subtends an angle at any point on it's alternate arc, that angle would be x⁄2
Is there any such point?
Indeed there is.
6. If we join A to N, we will see that, arc AXB subtends an ∠ANB at the point N on the alternate arc.
So ANB = x⁄2. This is shown in fig(d)
 |
| Fig.27.33 |
8. So, if this arc MYN subtends an angle at any point on it's alternate arc, that angle would be y⁄2
Is there any such point?
Indeed there is.
9. Arc MYN subtends an ∠MAN at the point A on the alternate arc.
So ∠MAN = y⁄2
10. Now consider PAN in fig(e).
• ∠ ANB is an exterior angle in ΔPAN
• Exterior angle = sum of remote interior angles.
11. So we get: ∠ANB = ∠APB + ∠PAN ⇒x⁄2 = ∠APB + x⁄2 ⇒ ∠APB = (x-y)⁄2
Theorem 27.6
1. Consider any arc AXB2. Let it subtend ∠APB at a point P in the exterior of the circle
3. Also let the two legs AP and BP intersect the circle at M and N respectively
4. Then ∠APB is calculated as follows:
(i) Write the central angle of arc AXB
(ii) Write the central angle of arc MYN
(iii) Subtract the smaller from the larger
(iv) Half of the result of subtraction will be ∠APB
Solved example 27.12
In the fig.27.34(a) below, central angle of arc AXB is 40o. Central angle of arc CYD is 70o.
Find the angles of ΔAPD.
 |
| Fig.27.34 |
Solution:
• P is an interior point. The arc AXB makes ∠APB at P
• The legs AP and BP are extended to meet the circle at C and D respectively.
• So we can use theorem 27.5 to calculate ∠APB. But in this problem we are not asked to find ∠APB
• We are asked to find the interior angles of ΔAPD. So direct application of theorem 27.5 is not possible
• We are asked to find the interior angles of ΔAPD. So direct application of theorem 27.5 is not possible
• However we need to apply some results that we came across while deriving theorem 27.5. The steps are given below:
1. Consider fig.27.34(b).
• Arc AXB and arc AYB are alternate arcs
• Arc AXB subtends ∠ADB on the alternate arc
• Arc AXB has a central angle of 40o. So ∠ADB = 1⁄2 × 40 = 20o
• But ∠ADB = ∠ADP. So we got the interior angle of ΔAPD at D
2. Consider fig.27.34(c).
• Arc CYD and arc CXD are alternate arcs
• Arc CYD subtends ∠DAC on the alternate arc
• Arc CYD has a central angle of 70o. So ∠DAC = 1⁄2 × 70 = 35o
• But ∠DAC = ∠DAP. So we got the second interior angle of ΔAPD at P
3. Now the third interior angle, which is at P = [180-(20+35)] = [180-55] = 125o.
4. So the required interior angles are:
∠D = 20o, ∠A = 35o and ∠P = 125o.
Solved example 27.13
In fig.27.35(a) below, AB is a diameter of the circle. A, P, B and R are four points on the circle. Lines AP and RB intersect at Q. Find ∠PRB, ∠PBR and ∠BPR
Solution:
1. Separate out arc PYB as shown in fig(b). It is subtending ∠BAP = 34o on the alternate arc PXB
2. This same arc PYB is subtending another ∠PRB also on the alternate arc PXB
So ∠BAP = ∠PRB = 34o.
3. Separate out arc AXR. It is subtending ∠ABR on the alternate arc AYR
So ∠ABR will be half of the central angle x of the arc AXR. That is., ∠ABR = x⁄2.
4. This ∠ABR is an exterior angle of ΔABQ
• Exterior angle = sum of remote interior angles
• x⁄2 = 34 + 28 ⇒ x⁄2 = 62
5. So from (3) above, we get: ∠ABR = x⁄2 = 62
6. The arc AXR is subtending ∠ABR on the alternate arc AYR
• The same arc AXR is subtending ∠APR on the alternate arc AYR
• So ∠ABR = ∠APR = 62o
7. Now consider ΔPRB. We are asked to find all it's interior angles
We have already obtained ∠PRB as 34o in (2)
8. AB is a diameter. So ∠APB = 90o. (Theorem 27.1)
∠BPR = ∠APB - ∠APR ⇒ ∠BPR = 90 - 62 ⇒ ∠BPR = 28o.
9. The remaining ∠PBR = [180 -(34+28)] = [180-62] = 118o.
10. So the required angles are:
∠PRB = 34o, ∠PBR = 118o and ∠BPR = 28o.
Solved example 27.13
In fig.27.35(a) below, AB is a diameter of the circle. A, P, B and R are four points on the circle. Lines AP and RB intersect at Q. Find ∠PRB, ∠PBR and ∠BPR
 |
| Fig.27.35 |
1. Separate out arc PYB as shown in fig(b). It is subtending ∠BAP = 34o on the alternate arc PXB
2. This same arc PYB is subtending another ∠PRB also on the alternate arc PXB
So ∠BAP = ∠PRB = 34o.
3. Separate out arc AXR. It is subtending ∠ABR on the alternate arc AYR
So ∠ABR will be half of the central angle x of the arc AXR. That is., ∠ABR = x⁄2.
4. This ∠ABR is an exterior angle of ΔABQ
• Exterior angle = sum of remote interior angles
• x⁄2 = 34 + 28 ⇒ x⁄2 = 62
5. So from (3) above, we get: ∠ABR = x⁄2 = 62
6. The arc AXR is subtending ∠ABR on the alternate arc AYR
• The same arc AXR is subtending ∠APR on the alternate arc AYR
• So ∠ABR = ∠APR = 62o
7. Now consider ΔPRB. We are asked to find all it's interior angles
We have already obtained ∠PRB as 34o in (2)
8. AB is a diameter. So ∠APB = 90o. (Theorem 27.1)
∠BPR = ∠APB - ∠APR ⇒ ∠BPR = 90 - 62 ⇒ ∠BPR = 28o.
9. The remaining ∠PBR = [180 -(34+28)] = [180-62] = 118o.
10. So the required angles are:
∠PRB = 34o, ∠PBR = 118o and ∠BPR = 28o.
In the next section, we will see Segments of a circle.
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