In the previous sections we saw theorem 27.4 and it's converse. We also saw some solved examples. In this section we will see a few more solved examples.
Solved example 27.8
How do we draw a 221⁄2o angle?
Solution:
• Let us first see what happens when we double a 221⁄2o angle. Fig.27.26 below shows a rough sketch
• In fig(a), an arc AB subtends an angle of 221⁄2o at P.
• Separate out this arc AB as shown in fig(b). We see that, the central angle of arc AB is (22.5 × 2) = 45o.
• Now we want to double this 45o. So we want this 45o in another circle. For that, with any convenient point O1 as centre and OO1 as radius, draw a circle. This is shown in fig(c)
• Since the centre is O1, and radius OO1, this circle will pass through O. So we have an angle of 45o in the new circle.
• This angle will cut an arc CD in the circle, and that arc CD will have a central ∠CO1D of 45 × 2 = 90o
• This 90o is shown in cyan colour in fig(c)
• So we made a 90o from a 221⁄2o.
■ We can do the reverse. That is., we can make a 221⁄2o from a 90o. The steps are as follows:
1. Draw a circle with centre O and also draw a diameter AB. This AB should preferably in the vertical direction, so that, our drawing will proceed towards the right in the horizontal direction. This is shown as green dashed line in fig.27.27(step 1) below.
2. The diameter AB splits the circle into two semi-circles. Mark a point O1 any where on the right semi-circle. Draw AO1 and BO1. By theorem 27.1, ∠AO1B will be 90o. This is shown in cyan colour in step 1. The point O1 should be approximately on a horizontal line through O. Then the construction will proceed towards the right in a horizontal direction.
3. Draw a circle with centre O1 and with any convenient radius so that, it will intersect both the legs of ∠AO1B. at points C and D. This is the red coloured circle in fig.27.27(step 2)
• So the arc CD have a central ∠CO1D of 90o.
4. Mark a point O2 on the red circle, approximately in line with O and O1. The angle subtended at O2 by the arc CD will be equal to (1⁄2×90) = 45o.
• So we halved 90o. We can continue like this and obtain the half of 45o.
5. Draw a circle with centre O2 and with any convenient radius so that, it will intersect both the legs of ∠CO2D. at points E and F. This is the green coloured circle in fig.27.27(step 3)
• So the arc EF has a central ∠EO2F of 45o.
6. Mark a point P on the green circle, approximately in line with O, O1 and O2. The angle subtended at P by the arc EF will be equal to (1⁄2×45) = 221⁄2o.
■ So we finally obtained 221⁄2o.
Solved example 27.9
In the fig,27.28(a) below, O is the centre of the circle and line OC is parallel to line PB.
(i) Prove that OC bisects ∠AOB
(ii) Explain how this can be used to draw the bisector of an angle
Solution:
Part (i):
1. Separate out the arc AB as shown in fig.27.28(b). This arc AB has a central ∠AOB of co.
2. Also, this arc AB subtends ∠APB at P on the alternate arc. Then by theorem 27.4, ∠APB = c⁄2o.
3. Given that OC is parallel to PB.
Consider fig(c):
• PB and OC are two parallel lines
• They are cut by a transversal PA
• So ∠PBA and ∠COA are corresponding angles and are equal
• Thus we get ∠COA = c⁄2o. That means, the line OC bisects ∠AOB
Part (ii):
1. Consider ∠AOB in fig.27.29(a) below. We want to draw the bisector of this angle
2. Draw a circle with centre O
3. Extend the line OA towards the left so that it meets the circle at P. This is shown in fig(b)
4. Draw PB
5. Draw OC parallel to PB. Then OC is the bisector of ∠AOB
■ Note that this problem is related to the special case that we saw before
Solved example 27.10
In the fig.27.30(a) below, O is the centre of the circle. Prove that x + y = 90o.
Solution:
1. Consider ΔAOB. The sides OA and OB are equal. Because they are radii of the same circle.
2. So ΔOAB is an isosceles triangle. The base angles will be equal. Thus ∠OAB = ∠OBA = xo. This is shown in fig(b)
3. So we get ∠AOB = [180 - (x+x)] = [180-2x]
4. Separate out the arc ADB. The ∠AOB is the central angle of arc ADB.
5. This arc ADB subtends an ∠ACB on the alternate arc
6. But ∠ACB is given as yo.
7. So we can write: ∠ACB = yo = half of [180-2x]
That is: y = [180-2x]⁄2 ⇒ y = (90-x) ⇒ (x+y) = 90o.
Solved example 27.11
In fig.27.31(a) below, PQ and RS are two mutually perpendicular chords of a circle.
∠QPR = 50o. Find ∠PQS
Solution:
1. Separate out the arc PR as shown in fig(b).
2. This arc PR subtends ∠PRS on the alternate arc
3. This same arc PR subtends another ∠PQS on the alternate angle
4. So ∠PRS = ∠PQS
5. We are asked to find ∠PQS. So if we can find ∠PRS, we get the answer.
6. Let the chords PQ and RS intersect at T
7. Consider ⊿PTR. It is given that PQ and RS are mutually perpendicular. So ∠PTR = 90o.
8. ∠PRS = [180-(90+50)] ⇒ ∠PRS = [180-140] ⇒ ∠PRS = 40o
9. Thus from (4) we get: ∠PQS = 40o.
Solved example 27.8
How do we draw a 221⁄2o angle?
Solution:
• Let us first see what happens when we double a 221⁄2o angle. Fig.27.26 below shows a rough sketch
• In fig(a), an arc AB subtends an angle of 221⁄2o at P.
• Separate out this arc AB as shown in fig(b). We see that, the central angle of arc AB is (22.5 × 2) = 45o.
• Now we want to double this 45o. So we want this 45o in another circle. For that, with any convenient point O1 as centre and OO1 as radius, draw a circle. This is shown in fig(c)
 |
| Fig.27.26 |
• This angle will cut an arc CD in the circle, and that arc CD will have a central ∠CO1D of 45 × 2 = 90o
• This 90o is shown in cyan colour in fig(c)
• So we made a 90o from a 221⁄2o.
■ We can do the reverse. That is., we can make a 221⁄2o from a 90o. The steps are as follows:
1. Draw a circle with centre O and also draw a diameter AB. This AB should preferably in the vertical direction, so that, our drawing will proceed towards the right in the horizontal direction. This is shown as green dashed line in fig.27.27(step 1) below.
2. The diameter AB splits the circle into two semi-circles. Mark a point O1 any where on the right semi-circle. Draw AO1 and BO1. By theorem 27.1, ∠AO1B will be 90o. This is shown in cyan colour in step 1. The point O1 should be approximately on a horizontal line through O. Then the construction will proceed towards the right in a horizontal direction.
 |
| Fig.27.27 |
• So the arc CD have a central ∠CO1D of 90o.
4. Mark a point O2 on the red circle, approximately in line with O and O1. The angle subtended at O2 by the arc CD will be equal to (1⁄2×90) = 45o.
• So we halved 90o. We can continue like this and obtain the half of 45o.
5. Draw a circle with centre O2 and with any convenient radius so that, it will intersect both the legs of ∠CO2D. at points E and F. This is the green coloured circle in fig.27.27(step 3)
• So the arc EF has a central ∠EO2F of 45o.
6. Mark a point P on the green circle, approximately in line with O, O1 and O2. The angle subtended at P by the arc EF will be equal to (1⁄2×45) = 221⁄2o.
■ So we finally obtained 221⁄2o.
Solved example 27.9
In the fig,27.28(a) below, O is the centre of the circle and line OC is parallel to line PB.
 |
| Fig.27.28 |
(ii) Explain how this can be used to draw the bisector of an angle
Solution:
Part (i):
1. Separate out the arc AB as shown in fig.27.28(b). This arc AB has a central ∠AOB of co.
2. Also, this arc AB subtends ∠APB at P on the alternate arc. Then by theorem 27.4, ∠APB = c⁄2o.
3. Given that OC is parallel to PB.
Consider fig(c):
• PB and OC are two parallel lines
• They are cut by a transversal PA
• So ∠PBA and ∠COA are corresponding angles and are equal
• Thus we get ∠COA = c⁄2o. That means, the line OC bisects ∠AOB
Part (ii):
1. Consider ∠AOB in fig.27.29(a) below. We want to draw the bisector of this angle
 |
| Fig.27.29 |
3. Extend the line OA towards the left so that it meets the circle at P. This is shown in fig(b)
4. Draw PB
5. Draw OC parallel to PB. Then OC is the bisector of ∠AOB
■ Note that this problem is related to the special case that we saw before
Solved example 27.10
In the fig.27.30(a) below, O is the centre of the circle. Prove that x + y = 90o.
 |
| Fig.27.30 |
1. Consider ΔAOB. The sides OA and OB are equal. Because they are radii of the same circle.
2. So ΔOAB is an isosceles triangle. The base angles will be equal. Thus ∠OAB = ∠OBA = xo. This is shown in fig(b)
3. So we get ∠AOB = [180 - (x+x)] = [180-2x]
4. Separate out the arc ADB. The ∠AOB is the central angle of arc ADB.
5. This arc ADB subtends an ∠ACB on the alternate arc
6. But ∠ACB is given as yo.
7. So we can write: ∠ACB = yo = half of [180-2x]
That is: y = [180-2x]⁄2 ⇒ y = (90-x) ⇒ (x+y) = 90o.
Solved example 27.11
In fig.27.31(a) below, PQ and RS are two mutually perpendicular chords of a circle.
 |
| Fig.27.31 |
Solution:
1. Separate out the arc PR as shown in fig(b).
2. This arc PR subtends ∠PRS on the alternate arc
3. This same arc PR subtends another ∠PQS on the alternate angle
4. So ∠PRS = ∠PQS
5. We are asked to find ∠PQS. So if we can find ∠PRS, we get the answer.
6. Let the chords PQ and RS intersect at T
7. Consider ⊿PTR. It is given that PQ and RS are mutually perpendicular. So ∠PTR = 90o.
8. ∠PRS = [180-(90+50)] ⇒ ∠PRS = [180-140] ⇒ ∠PRS = 40o
9. Thus from (4) we get: ∠PQS = 40o.
A solved example is shown in the form of a video presentation here:
Solved example video presentation
Solved example video presentation
In the next section, we will see the cases when an arc subtends an angle at a point P which is not on the circle.
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