Chapter 32.4 - Solved examples on Tangents and Chords

In the previous section we saw the relations between tangents and chords. In this section we will see some solved examples based on that discussion.

Solved example 32.11
In fig.32.33 (a) below, the circumcircle of ΔPQR is shown in yellow colour. ∠PQR = 80^o and ∠PRQ = 60^o. 

Fig.30.33

Red tangents are drawn at P, Q and R. The tangents intersect at 3 points to form ΔABC. Find the angles of ΔABC.
Solution:
• Given that ∠PQR = 80^oand ∠PRQ = 60^o. 
    ♦ So ∠RPQ = [180 -(80+60)] = 40^o
1. Consider the side PR. It is a chord of the yellow circle. 
• This chord subtends an angle of 80^o on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be 80^o
• That is., ∠BPR = ∠BRP = 80^o. 
• So, in ΔPRB, ∠B =  [180 -(80+80)] = 20^o. This is shown in fig.(b)
2. Consider the side QR. It is a chord of the yellow circle. 
• This chord subtends an angle of 40^o on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be 40^o
• That is., ∠QRA = ∠AQR = 40^o. 
• So, in ΔAQR, ∠A =  [180 -(40+40)] = 100^o.
3. Consider the side PQ. It is a chord of the yellow circle. 
• This chord subtends an angle of 60^o on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be 60^o
• That is., ∠CPQ = ∠CQP = 60^o. 
• So, in ΔCQP, ∠C =  [180 -(60+60)] = 60^o.

Solved example 32.12
In fig.32.34 (a) below, the circumcircle of ΔPQR is shown in yellow colour.  

Fig.32.34

Red tangents are drawn at P, Q and R. The tangents intersect at 3 points to form ΔABC. 
∠ABC = 40^o and ∠ACB = 60^o. Find the angles of ΔPQR.
Solution:
• Given that ∠ABC = 40^o and ∠ACB = 60^o.
    ♦ So ∠BAC = [180 -(40+60)] = 80^o
1. Consider ΔPRB. It is an isosceles triangle (∵ PB = RB. See Theorem 32.2)
• So base angles will be equal. We can write:
(2 × ∠BPR) + 40 = 180 ⟹ ∠BPR = ∠PRB = 70^o. This is shown in fig(b)
• Consider the side PR. It is a chord of the yellow circle. 
• The angles (at the two ends of that chord) on the tangent side is 70^o
• So this chord will subtend an angle of 70^o on the non-tangent side
• That is., ∠PQR = 70^o.
2. Consider ΔQRA. It is an isosceles triangle (∵ QA = RA. See Theorem 32.2)
• So base angles will be equal. We can write:
(2 × ∠AQR) + 80 = 180 ⟹ ∠AQR = ∠ARQ = 50^o. This is shown in fig(b)
• Consider the side QR. It is a chord of the yellow circle. 
• The angles (at the two ends of that chord) on the tangent side is 50^o
• So this chord will subtend an angle of 50^o on the non-tangent side
• That is., ∠QPR = 50^o.
3. Consider ΔQPC. It is an isosceles triangle (∵ QC = PC. See Theorem 32.2)
• So base angles will be equal. We can write:
(2 × ∠CQP) + 60 = 180 ⟹ ∠CQP = ∠QPC = 60^o. This is shown in fig(b)
• Consider the side QP. It is a chord of the yellow circle. 
• The angles (at the two ends of that chord) on the tangent side is 60^o
• So this chord will subtend an angle of 60^o on the non-tangent side
• That is., ∠QRP = 60^o.

Solved example 32.13
In the fig.32.35(a) given below, PQ, RS and TU are tangents to the circumcircle of ABC. Sort out the equal angles in the fig.

Fig.32.35

Solution:
1. Consider the side AC. It is a chord of the yellow circle. 
• This chord subtends ∠ABC on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be equal to ∠ABC
• That is., ∠PAC = ∠TCA = ∠ABC. These are shown in green color in fig(b).
2. Consider the side BC. It is a chord of the yellow circle. 
• This chord subtends ∠BAC on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be equal to ∠BAC
• That is., ∠UCB = ∠RBC = ∠BAC. These are shown in magenta color.
3. Consider the side AB. It is a chord of the yellow circle. 
• This chord subtends ∠ACB on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will be equal to ∠ACB
• That is., ∠QAB = ∠SBA = ∠ACB. These are shown in cyan color.

Solved example 32.14
In the fig.32.36(a) below, O is the center of the circle and AB is the tangent to the circle through Q. Find ∠PQA

Fig.32.36

Solution:
1. Given that, the central angle of chord PQ is 100^o. So ∠PRQ = 50^o. [Theorem 27.4]
• This is shown in fig(b).
2. In ΔPQR, consider the side PQ. It is a chord of the yellow circle. 
• This chord subtends an angle of 50^o on the non-tangent side
• So the angles (at the two ends of that chord) on the tangent side will also be 50^o
• That is., ∠PQA = ∠PRQ = 50^o.

Solved example 32.15
In the fig.32.37(a) below, O is the center of the circle and QR is a diameter of the circle through Q. AB is a tangent to the circle at P. If ∠BPQ = 50^o, find ∠PQR

Fig.32.37

Solution:
1. In ΔPQR, consider the side PQ. It is a chord of the yellow circle
• The tangent AB is at the end P of the chord
• Draw a tangent CD at the end Q also. This is shown in fig(b)
2. The angle at the end P on the tangent side is given to be 50^o
• So the angle at end Q on the tangent side will also be 50^o
• That is., ∠DQP = 50^o
3. But ∠DQO = 90^o (∵ the tangent CD is at the end Q of the diameter QR)
4. So we get: ∠PQR = (∠DQO - ∠DQP) = (90 - 50) = 40^o

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In the next section, we will see more details about Tangents.

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Chapter 32.3 - Tangent and Chord

In the previous section we saw tangents giving cyclic quadrilaterals. In this section we will learn the relations between tangents and chords.

1. A circle is drawn with center at O. See fig.32.28(a)

Fig.32.28

2. Two green radial lines OP and OQ are drawn in such a way that, ∠POQ = 100
3. A red tangent is drawn at P. Another red tangent is drawn at Q
4. The red tangents meet at T. So the tangents are named as AT and BT
5. A magenta line is drawn joining P and Q. So the magenta line is a chord
• Clearly, the central angle of the chord is 100^o  
6. We know that ∠PTQ will be equal to 80^o [Theorem 32.5]
• This is shown in fig(b)
7. We also know that PT = QT. [Theorem 32.2]
8. So ΔPTQ is an isosceles triangle. It's base angles are equal. Let them be x
9. So in ΔPTQ, we get: (2x + 80) = 180^o ⟹ x = 50^o
10. But this 50^o is half of the central angle (made by the chord PQ) 100^o 

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Let us see if this is true for any chord:
1. A circle is drawn with center at O. See fig.32.29(a) 

Fig.32.29

2. Two green radial lines OP and OQ are drawn in such a way that, ∠POQ = c^o
3. A red tangent is drawn at P. Another red tangent is drawn at Q
4. The red tangents meet at T. So the tangents are named as AT and BT
5. A magenta line is drawn joining P and Q. So the magenta line is a chord
• Clearly, the central angle of the chord is c^o
6. We know that ∠PTQ will be equal to (180-c)^o  [Theorem 32.5]
7. We also know that PT = QT. [Theorem 32.2]
8. So ΔPTQ is an isosceles triangle. It's base angles are equal. Let them be x^o
9. So in ΔPTQ, we get: [2x + (180-c)] = 180^o ⟹ (2x - c)  = 0 ⟹ x = (^c⁄₂)^o

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We can write the above result as a theorem:
Theorem 32.6
• P and Q are two points on the circle
• A tangent is drawn at P
    ♦ The angle between this tangent and the chord PQ is half of the central angle of the chord
• A tangent is drawn at Q
    ♦ The angle between this tangent and the chord PQ is also half of the central angle of the chord

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Now we will see a more interesting case:
1. The chord PQ in fig.32.29 above, divides the circle into two arcs: 
• A minor arc PQ and a major arc PQ
2. Consider any point R on the major arc PQ. This is shown in fig.32.30(a)

Fig.32.30

• The chord PQ will subtend  ∠PRQ on that major arc. 
3. We know that ∠PRQ will be equal to half the central angle of the chord PQ. [Theorem 27.4] 
• So we get: ∠PRQ = (^c⁄₂)^o
4. But based on the theorem 32.6 that we wrote above, the angle between the chord and the tangent is also (^c⁄₂)^o
5. Thus we get an interesting result: ∠PRQ = ∠TPQ = ∠TQP

Let us see some examples:
1. In fig.32.30(b), PQ is a chord. 
• Tangent is drawn at P
• Tangent is drawn at Q
• The two tangents meet at T
2. The meeting point T is on the right side of the chord
• We will call the right side as 'Tangent side of the chord'
• So the left side of the chord can be called: 'Non-tangent side of the chord'
3. A point R is marked on the circle on the 'Non-tangent side of the chord'
■ Wherever we mark R on the major arc, the ∠PRQ will be the same. [Details here]
• In our present case, it is 65^o. 
4. Then, from what we have seen based on fig.32.30(a), we can directly write:
■ The angle between the chord and tangent will be 65^o at both ends of the chord
• That is., ∠TPQ and ∠TQP will be equal to 65
• We do not need to know the position of the center 'O' for writing the above result
5. Note that, ∠TPQ and ∠TQP are the angles on the Tangent side of the chord PQ
• Which are the angles on the Non-tangent side?
They are: ∠APQ and ∠BQP
6. What are their values?
• We know that ∠APQ and ∠TPQ form a linear pair
• So ∠APQ = (180 - ∠TPQ) = (180 - 65) = 115^o. This is shown in fig (c) 
• Similarly, ∠BQP = 1(80 - ∠TQP) = (180 - 65) = 115^o
7. Now consider any point S on the circle on the tangent side of the chord
• PRQS is a cyclic quadrilateral. So ∠PSQ = (180 - ∠PRQ) = (180 - 65) = 115^o. This is shown in fig(c)
• Note that, in figs.(b) and (c), we do not know where the center of the circle is. Even then we are able to find various angles

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We will write the above findings as a theorem:
Theorem 32.7
1. A chord PQ subtends ∠PSQ  at a point S on the circle, on the Tangent side
• At the end P, the chord makes ∠APQ (with the tangent AT) on the Non-tangent side 
• At the end Q, the chord makes ∠BQP (with the tangent BT) on the Non-tangent side 
■ All the above three angles are equal
2. The chord PQ subtends ∠PRQ  at a point R on the Non-tangent side
• At the end P, the chord makes ∠QPT (with the tangent AT) on the Tangent side 
• At the end Q, the chord makes ∠TQP (with the tangent BT) on the Tangent side 
■ All the above three angles are equal
• In (1), point S is on the Tangent side. The chord angles are on the Non-tangent side
• In (2), point R is on the Non-tangent side. The chord angles are on the Tangent side

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Now we will see the practical application of the above theorem 32.7
We know how to draw the tangent at a given point on a circle. 
1. Draw a radial line through the given point
2. Draw a line perpendicular to the radial line through the given point. This line is the required tangent.
• So the procedure is simple. But it will not be so simple if the centre of the circle is not known. Because, we will not be able to draw the radial line.
• In such a situation, we can use the above theorem 32.7. Let us see the method:
• In fig.32.31(a) below, a circle is shown and a point P is marked on it. Center of the circle is not given. We are required to draw the tangent at P. 

Fig.32.31

We can use the following procedure:
1. From the point P, draw a convenient chord PQ. This is shown in fig(b)
2. Mark a convenient point R on the circle on the Non-tangent side. Complete the triangle PQR
3. Measure ∠PRQ. Let it be x^o
4. Draw a red line through P in such a way that the angle (at P) between the red line and the chord is x^o. This is shown in fig(c)
5. Then the red line is the required tangent
An easier method:
• In any case, we will need to draw PQR. This is to obtain the value of x
1. We will draw PQR in such a way that, it is isosceles. For that, with P as centre, draw two arcs cutting the circle at Q and R. This is shown in fig.32.32(a) below:

Fig.30.32

2. Now complete ΔPQR. It is an isosceles triangle because PR = PQ. 
• The base angles at R and Q will be the same.
3. Draw a red line through P in such a way that it is parallel to RQ. Then we have:
• RQ and the red line are two parallel lines. They are cut by a transversal PQ
• So the following two angles are co-interior angles and hence will be equal:
    ♦ ∠RQP
    ♦ The angle (at P) between the red line and PQ
4. So we get:
• ∠PRQ = x = ∠RQP = The angle (at P) between the red line and PQ
• So the angle at P is also x. This is shown in fig c. 
■ Thus the red line is the tangent at P

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An actual construction can be seen in the form of a video presentation here.

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In the next section, we will see some solved examples.

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Chapter 30.6 - Trigonometry related to Circles

In the previous section we saw some solved examples on 30^o, 60^o and 45^o right triangles. In this section, we will see some applications of trigonometry related to circles.

• Consider the arc ABC in fig.30.23(a) below. It has a central angle of θ^o. Radius of the circle is 'r' cm.

• Line segment AC which joins the ends of the arc ABC is a chord of the circle.
• We have seen how to calculate the length of the arc ABC in previous classes. See details here.
• We have also seen how to calculate the length of chord AC. See details here.  In this section we will see another method using Trigonometry.
1. Consider the ΔAOC in fig.30.23(b). Drop a perpendicular OD from the vertex O onto the side AC. This perpendicular will bisect AC. So AD = CD. The reason can be seen here.
2. Also, the perpendicular OD will bisect ∠AOC. So ∠AOD = ∠COD = ^θ⁄₂ .
3. Now we have two right triangles: ⊿AOD and ⊿COD. Consider any one of them. Let us consider ⊿AOD. We get: sin ^θ⁄₂  = ^AD⁄_AO = ^AD⁄_r  ⟹ AD = r × sin ^θ⁄₂. = r sin ^θ⁄₂.
4. But AD = CD. That is., AD is half of the chord AC
• So the full length of the chord AC = 2 × r sin ^θ⁄₂. = 2r sin ^θ⁄₂.
• In this way we can find the length of any chord if the central angle and the radius of circle are known. We can write it in the form of a theorem. We will write it in steps:
Theorem 30.1:
1. Consider any chord in a circle of radius r
2. Let the chord make a central angle of ^θ⁄₂
3. Find the sine of ^θ⁄₂
4. Multiply this sine by 2r
5. The product will be the length of the chord.
That is., Length of the chord = 2r sin ^θ⁄₂

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An example:
Find the length of a chord whose central angle is 70. Radius of the circle is 3 cm
Solution:
1. Given: r = 3 cm, θ = 70^o
2. Then length of the chord = 2r sin ^θ⁄₂ = 2 × 3 × sin ⁷⁰⁄₂ = 6 sin 35 = 6 × 0.5736 = 3.4414 cm    
• A diagram of this problem drawn with a computer program gives the same result. It is shown below:

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Another application:

• Consider any three points on a plane sheet of paper. If those three points are not on a line, we can draw a circle passing through all those three points. 
• Also, a triangle can be formed using those three points. That means, we can draw a circle passing through the three vertices of any given triangle. It is called the circumcircle of the triangle. We learned all those details here.
• Consider the triangle and it's circle shown in the fig.30.24(a) below:

• The angle at vertex A is given. The radius of the circle is also given. We want to find the length of the chord BC. Is it possible? Let us try:
1. Consider the minor arc BC in fig.30.24(b). Our required chord joins the two end points of this minor arc BC
2. The minor arc BC makes an angle A^o on the alternate arc BAC. So the central angle of minor arc BC will be (2A)^o. Details here. 
3. So the central angle of the chord BC is also (2A)^o.
4. Once we get the central angle of chord BC, we can easily find it's length. For that, we use theorem 30.1 that we saw above.
• Length of chord BC = 2r sin ^θ⁄₂ = 2r sin ^(2A)⁄₂ = 2r sin A 
5. Note that, BC is the opposite side of vertex A. If we use another vertex B, we will get the length of chord AC. That is.,
• Length  of chord AC = 2r sin B 
• Similarly, Length  of chord AB = 2r sin C

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Now we have to check whether this formula is applicable to all triangles.   

1. Consider the circumcircle of ΔABC in fig.30.25(a) below. 

•The centre of the circle is outside ΔABC. Note that, in the previous fig.30.24(a), the centre is inside ΔABC

2. The arc BC is now a major arc. This is shown in red colour in fig(b). The central angle of this arc is (2A)^o

3. Now consider fig(c). We have, ∠COB = (360-2A)

4. Drop a perpendicular OD from O onto the side BC. This will bisect BC and also ∠COB

5. So we get: ∠COD = ∠BOD = ^COB⁄₂  = ^(360-2A)⁄₂ = ^(180-A)⁄₂ 

6. In ⊿OCD, sin ^(180-A)⁄₂ = ^CD⁄_r ⟹ CD = r sin ^(180-A)⁄₂ 

7. So BC = 2 × CD = 2r sin ^(180-A)⁄₂.   

• So we can write:
If an angle of a triangle is obtuse, we must subtract it from 180, to use in the formula.

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An example:

In fig.30.26 below, length of side AB is given. Find the lengths of the other two sides

Solution:

We have: AB = 2r sin C ⟹ 6 = 2r sin 80 ⟹ 6 = 2r × 0.9848 ⟹ 2r = ⁶⁄_0.9848 

⟹ 2r = 6.0926 cm

2. Now, BC = 2r sin A

Substituting the values of '2r' and 'A', we get: BC = 6.0926 × sin 60 

⟹ BC = 6.0926 × 0.8660 = 5.28 cm

3. Similarly, AC  = 2r sin B

Substituting the values of '2r' and 'B', we get: AC = 6.0926 × sin 40 

⟹ BC = 6.0926 × 0.6428 = 3.92 cm

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Using the sine and cosine tables, and if needed, a calculator, do the following problems:

Solved example 3.16

A triangle and it's circumcircle are shown in the fig.30.27(a) below. Calculate the diameter of the circle

Solution:
1. Let us name the triangle as ΔABC. This is shown in fig(b). Let the radius of the circumcircle be 'r' cm
2. Then we get: AB = 2r sin C ⟹ 4 = 2r × sin 70 ⟹ 4 = 2r × 0.9397 ⟹ 2r = ⁴⁄_0.9397 = 4.26 cm
3. But 2r is the diameter. So we can write:
• Diameter of the circumcircle = 4.26 cm

Solved example 3.17
A circle is to be drawn, passing through the ends of a line 5 cm long. This line should subtend an angle of 80^o on one side. What should be the radius of the circle?
Solution:
Consider the rough sketch in fig.30.28 below

1. Let us name the triangle as ΔABC. Let the radius of the circumcircle be 'r' cm
2. Then we get: AB = 2r sin C ⟹ 5 = 2r × sin 80 ⟹ 5 = 2r × 0.9848
⟹ 2.5 = r × 0.9848 ⟹ r = ^2.5⁄_0.9848 = 2.538 cm
3. So we can write:
• Radius of the circumcircle = 2.538 cm

Solved example 30.18
A part of a circle is shown in fig.30.29(a) below. What is the radius of the circle?

Solution:
1. The possible full circle is shown in fig(b). We have to find the radius 'r' of this circle
2. Consider the minor arc ADC. It subtends ∠ABC on the alternate arc. Now we have a cyclic quadrilateral ABCD
3. In the cyclic quadrilateral, ∠ADC + ∠ABC = 180^o. So ∠ABC = 180 - 140 = 40^o
4. Consider ΔABC. We get: AC = 2r sin B ⟹ 8 = 2r sin 40 ⟹ 8 = 2r × 0.6428
⟹ 2r = ⁸⁄_0.6428 ⟹ r = ⁴⁄_0.6428 = 6.222 cm

Another method:
1. In fig(b), ∠ADC is an obtuse angle. So, to relate it with the length of chord AC, we must subtract it from 180^o. (see fig.30.25 above)
2. We get: AC = 2r sin (180-140) ⟹ 8 = 2r sin 40 ⟹ 8 = 2r × 0.6428
⟹ 2r = ⁸⁄_0.6428 ⟹ r = ⁴⁄_0.6428 = 6.222 cm

Solved example 30.19
Draw the circle and the triangle shown in fig.30.30(a) in your note book and explain how it was drawn. Calculate the lengths of all three sides

Solution:
1. In fig(a) we have a circle of diameter 5 cm, and a triangle whose two angles are given
2. The third angle will be obviously [180-(45+65)] = [180 - 110] = 70^o.
3. The given triangle is named as ΔABC in fig(b). Two more details are also added in this fig(b):
• The radius of the circle is 2.5 cm
• The minor arc AB subtends an angle of 70^o on the alternate arc. So the central angle ∠AOB of this minor arc will be equal to 70 × 2 = 140^o
• With these details we can do the construction. The steps are shown in the fig.30.31 below:

Step 1: • With any convenient point 'O' as centre, draw a circle of radius 2.5 cm
• At 'O', draw two lines with an angle of 140^o between them
• Name the points of intersection of the lines with the circle as 'A' and 'B'
• These are shown in fig.30.31(a) above
Step 2: • Draw line AB. This is shown in fig(b)
Step 3: • At A, draw a second line at an angle of 45^o with AB
• At B, draw a third line at an angle of 65^o with AB
• These two lines will intersect at a point. This point will lie on the circle. Name this point as 'C'.
• Thus the construction is complete
Part 2: In this part, we have to calculate the sides of the ΔABC
1. AB = 2r sin C = 2×2.5×sin 70 = 2×2.5×0.9397 = 4.6985 cm
2. AC = 2r sin B = 2×2.5×sin 65 = 2×2.5×0.9063 = 4.5315 cm
3. BC = 2r sin A = 2×2.5×sin 45 = 2×2.5×0.7071 = 3.5355 cm

Solved example 30.20
A triangle is made by drawing angles of 50^o and 65^o at the ends of a 5 cm long line. Calculate it's area
Solution:
The given data is shown in fig.30.32(a) below:

1. Let us add some more details. The triangle is named as ΔABC. The third angle at C will be [180 - (50+65)] = [180-115] = 65^o. This is shown in fig(b)
2. So two angles are 65^o. It is an isosceles triangle. Sides opposite the equal angles are equal. 
We get: AB = AC = 5 cm
3. Drop a perpendicular CD from the vertex C to the side AB. Consider the right triangle ⊿ADC.
We have: sin 50 = ^CD⁄_AC ⟹ 0.7660 = ^CD⁄₅ ⟹ CD = 0.7660 × 5 = 3.83 cm
4. Area of ΔABC = ¹⁄₂ × base × altitude = ¹⁄₂ × 5 × 3.83 = 9.58 cm²

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In the next section we will see problems related to the other trigonometric ratio tan.

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Chapter 27.8 - Cyclic quadrilaterals - Solved examples

In the previous section we saw theorem 27.9 and it's converse. In this section we will see some solved examples.

Solved example 27.15
In the fig.27.46(a) below, A, B, C and D are four points on a circle. 

Fig.27.45

∠DBC = 55^o and ∠CAB = 45^o. Compute ∠BCD
Solution:

1. Consider arc BC in fig(b). It subtends ∠BAC (= 45^o) on the alternate arc. 

• The same arc subtends ∠BDC on the alternate arc. So ∠BDC = ∠BAC = 45^o. This is marked in fig(b)
2. Consider ΔBDC. We get ∠BCD = [180-(45+55)] =[180-100] = 80^o. (∵ sum of interior angles of a triangle is 180^o)
Thus we get the required angle. We can do a check by using theorem 27.9.   

3. Consider arc CD in fig(c). It subtends ∠CBD (= 55^o) on the alternate arc. 

• The same arc subtends ∠CAD on the alternate arc. So ∠CBD = ∠CAD = 55^o. This is marked in fig(c)
(i) Sum of opposite angles ∠BAD and ∠BCD = 45 + 55 + 80 = 180^o.          

Solved examples 27.16
In the fig.27.46(a) below, A, B, C and D are four points on a circle. 

Fig.27.46

AC and BD intersect at E in such a way that ∠BEC = 30^o. and ∠ECD = 20^o. Find ∠BAC
Solution:
1. ∠BEC and ∠BEA form a linear pair. So ∠BEA = 180 - ∠BEC = 180 -130 = 50^o.
2. ∠BEA and ∠CED are opposite angles, and are hence equal. So we get ∠BEA = ∠CED = 50^o.

3. Consider ΔCED. We get ∠EDC = [180-(50+20)] =[180-70] = 110^o. (∵ sum of interior angles of a triangle is 180^o)

4. Consider the major arc BC in fig(c). It subtends ∠BDC (= 110^o) on the alternate arc. 

• The same arc subtends ∠BAC on the alternate arc. So ∠BAC = ∠BDC = 110^o. This is marked in fig(c). Thus we get the required angle.

Solved example 27.17
In the fig.27.47(a) below, ABCD is a square. 

Fig.27.47

Determine ∠DPC
Solution: 
1. Draw the diagonal AC of the square. 
2. A diagonal of a square will bisect the angles at the corners. So we get:
∠DAC = ∠BAC = 45^o.
3. Consider the quadrilateral ACPD. It is a cyclic quadrilateral. The sum of opposite angles = 180^o.
4. So we get: ∠DPC + ∠DAC = 180^o ⇒ ∠DPC + 45 = 180 ⇒ ∠DPC = 180 - 45 = 135^o.

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Now we will see an important result related to cyclic quadrilaterals. We will learn it in steps: 

1. Fig.27.48(a) below shows a cyclic quadrilateral ABCD. 

Fig.27.48

2. The side AB is extended along towards the right up to point E. So ∠CBE (shown in red colour) becomes an exterior angle of the cyclic quadrilateral. We can write:
■ ∠CBE is the exterior angle of the cyclic quadrilateral ABCD at the vertex B. 
3. For the vertex B, the opposite vertex is D
• So, for the vertex B, ∠ADC (shown in yellow colour) is the 'interior angle at the opposite vertex'
4. Thus we have three quantities:
(i) A vertex B  (ii) Exterior angle at that vertex B  (iii) Interior angle at D, which is the opposite vertex of B  
• We want to know the relation between (ii) and (iii)
5. Consider the interior angle at B. It is shown in blue colour in fig (b)
• Blue + Red will obviously be 180^o (∵ they form a linear pair)
• So we can write: ∠CBE + ∠ABC = 180^o
6. Yellow and Blue are opposite angles of a cyclic quadrilateral. So their sum will be 180^o. 
• We can write: ∠ADC + ∠ABC = 180^o
7. From (5) we get: ∠ABC = 180 – ∠CBE
• Substituting this in (6) we get:
∠ADC + (180 – ∠CBE) = 180
⇒ ∠ADC – ∠CBE = 180 – 180  
⇒ ∠ADC – ∠CBE = 0 
⇒ ∠ADC = ∠CBE
8. So we can write: 
• The exterior angle at B is equal to the interior angle at opposite vertex
9. Now consider fig (c). The side CB is extended upto F
• ∠ABF is an exterior angle at vertex B. So is ∠CBE
• But we can see that the two are equal because, they are opposite angles. 
• So, at a vertex, there will be only one value for an exterior angle
10. We can write the above results in a general form:
• Consider any vertex of a cyclic quadrilateral. 
• There will be an exterior angle at that vertex
• That exterior angle will be equal to the interior angle at the opposite vertex

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Some solved examples on cyclic quadrilaterals are shown in the form of a video presentation at the following links:
Trapezium Cyclic or not

Non-rectangular parallelogram Cyclic or not

Video presentation 1 

Video presentation 2 

Video presentation 3 

Video presentation 4 

Parallelogram with one vertex at center

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In the next section, we will learn about Multiplication of Chords.

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