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 27.9 - Chords inside a Circle

In the previous section we completed the discussion on cyclic quadrilaterals. In this section we will see chords. We saw some basic details about chords in chapter 17.3. Now we will see some advanced details.

• Consider any two chords of a circle. Only condition is that, they must be non-parallel. 
• Since they are non-parallel, they will surely intersect at a point 'P'. 
• This 'P' may be inside the circle as shown in fig.27.49(a). Or 'P' may be outside the circle as shown in fig.27.49(b).

Fig.27.49

■ Which ever be the case, there are some similarities between the two. Let us analyse:
1. Consider fig.27.50(a) below. It is the same fig.27.49(a) that we saw above. 

Fig.27.50

2. A small modification is made. That is., AD and BC are drawn with red lines. That is., ends  of one chord are joined to the ends of the other. 
3. Now we get two triangles: ΔAPD and ΔBPC. 
• These two triangles are similar. We can prove this as follows:
4. In the fig.27.51(a) below, a chord BD is drawn. 

Fig.27.51

5. This chord BD divides the circle into two segments. Take out the larger segment. 
6. ∠DAB and ∠DCB are two angles in this larger segment. So they are both equal to the unique angle. See theorem 27.7. That means both have the same angle value. 
7. So we can write this:    
• ∠DAP in ΔAPD is equal to ∠BCP in ΔBPC
• In other word, ∠A in ΔAPD is equal to ∠C in ΔBPC (They are shown in yellow colour)
8. Let us see if there are any other equal angles like them:
• Consider ∠APD and ∠BPC. They are opposite angles, and hence equal. So we can write:
• ∠P in ΔAPD is equal to ∠P in ΔBPC (They are shown in white colour)
9. Thus we get 'two angles the same' in the two triangles ΔAPD and ΔBPC. 
• Naturally the third angle must also be the same. 
However, we will write the calculation steps:
(i) The third angle (ie., ∠D) in ΔAPD = 180 - ∠A - ∠P 
(ii) The third angle (ie., ∠B) in ΔBPC = 180 - ∠C - ∠P 
(iii) But ∠C = ∠A and ∠P = ∠P. So (i) will be equal to (ii). That is., ∠D will be equal to ∠B
10. Thus all the angles in the two triangles ΔAPD and ΔBPC are equal
■ So they are similar triangles
11. Now we apply a special property that is applicable to any two similar triangles:
^{side opposite smallest angle in ΔAPD}⁄_{side opposite smallest angle in ΔBPC} 
= ^{side opposite medium angle in ΔAPD}⁄_{side opposite medium angle in ΔBPC} 
= ^{side opposite largest angle in ΔAPD}⁄_{side opposite largest angle in ΔBPC}
12. But angles in the two triangles are the same. That is.,
• Smallest angle in ΔAPD = Smallest angle in ΔBPC
• Medium angle in ΔAPD = Medium angle in ΔBPC
• Largest angle in ΔAPD = Largest angle in ΔBPC
13. So we can write this:
• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,
^{side opposite ∠A in ΔAPD}⁄_{side opposite ∠C in ΔBPC} 
= ^{side opposite ∠D in ΔAPD}⁄_{side opposite ∠B in ΔBPC}
= ^{side opposite ∠P in ΔAPD}⁄_{side opposite ∠P in ΔBPC}
14. So we get: ^PD⁄_PB = ^AP⁄_PC = ^AD⁄_BC .

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Now let us see if we can derive the same result in (14) for the case when P is outside the circle:

1. Consider fig.27.50(b) above. It is the same fig.27.49(b) that we saw earlier. 
2. A small modification is made. That is., AD and BC are drawn with red lines. That is., ends  of one chord are joined to the ends of the other. 
3. Now we get two triangles: ΔAPD and ΔBPC. [Note that, even though 'P' is outside the circle, we get triangles with the same names as in the previous case] 
• These two triangles are similar. We can prove this as follows:
4. In the fig.27.51(b) above, a chord BD is drawn.
5. This chord BD divides the circle into two segments. Take out the larger segment. 
6. ∠DAB and ∠DCB are two angles in this larger segment. So they are both equal to the unique angle. See theorem 27.7. That means both have the same angle value. 
7. So we can write this:    
• ∠DAP in ΔAPD is equal to ∠BCP in ΔBPC
• In other word, ∠A in ΔAPD is equal to ∠C in ΔBPC (They are shown in yellow colour)
8. Let us see if there are any other equal angles like them:
• Consider ∠APD and ∠BPC. They are one and the same, and hence equal. So we can write:
• ∠P in ΔAPD is equal to ∠P in ΔBPC (This is shown in white colour)
9. Thus we get 'two angles the same' in the two triangles ΔAPD and ΔBPC. 
• Naturally the third angle must also be the same. 
However, we will write the calculation steps:
(i) The third angle (ie., ∠D) in ΔAPD = 180 - ∠A - ∠P 
(ii) The third angle (ie., ∠B) in ΔBPC = 180 - ∠C - ∠P 
(iii) But ∠C = ∠A and ∠P = ∠P. So (i) will be equal to (ii). That is., ∠D will be equal to ∠B
10. Thus all the angles in the two triangles ΔAPD and ΔBPC are equal
■ So they are similar triangles
11. Now we apply a special property that is applicable to any two similar triangles:
^{side opposite smallest angle in ΔAPD}⁄_{side opposite smallest angle in ΔBPC} 
= ^{side opposite medium angle in ΔAPD}⁄_{side opposite medium angle in ΔBPC} 
= ^{side opposite largest angle in ΔAPD}⁄_{side opposite largest angle in ΔBPC}
12. But angles in the two triangles are the same. That is.,
• Smallest angle in ΔAPD = Smallest angle in ΔBPC
• Medium angle in ΔAPD = Medium angle in ΔBPC
• Largest angle in ΔAPD = Largest angle in ΔBPC
13. So we can write this:
• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,
^{side opposite ∠A in ΔAPD}⁄_{side opposite ∠C in ΔBPC} 
= ^{side opposite ∠D in ΔAPD}⁄_{side opposite ∠B in ΔBPC}
= ^{side opposite ∠P in ΔAPD}⁄_{side opposite ∠P in ΔBPC}
14. So we get: ^PD⁄_PB = ^AP⁄_PC = ^AD⁄_BC.

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So we get the same result (14) in both the cases. That means, the result is valid for both 'P inside' and 'P outside' the circle.
1. Now, from among the three ratios, we will take out two, which has 'P'. So we take out the first and second. So we get:
^PD⁄_PB = ^AP⁄_PC  
2. Cross multiplying we get: PA × PB = PC × PD

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So we can always multiply the opposite pieces. We will write this result as a theorem.

Theorem 27.10:

1. Two chords of a circle meet at a point inside the circle

2. The point divides the each chord into two pieces

3. Multiply the two pieces belonging to one chord

4. Multiply the two pieces belonging to the other chord

5. The two products will always be equal

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Let us now see one application of the above theorem. We will see it as a solved example:

Solved example 27.18
The distance between the ends of a piece of bangle is 4 cm. It’s height is 1 cm. What is the radius of the full bangle?
Solution:
We did this problem when we learned about length of chords. See here. Now we will do it using another method:
1. In the fig.27.53(a) below, A and B are the ends of the bangle. The distance between them is 4 cm.

Fig.27.53

2. The height of the piece is given as 1 cm. It should be measured in a direction perpendicular to the line AB. This is also shown in fig.27.53(a)
3. In fig(b), the remaining portion of the bangle is shown in dashed line.  
4. AB is a chord of the full circle. 
5. Consider a diameter CD. It must satisfy one condition:
It must be perpendicular to the chord AB
6. If this condition is satisfied, the diameter CD will bisect the chord AB at P. See Theorem 17.1.
7. When AB is bisected, AP = BP = 2 cm
8. Since the diameter is also a chord, we can apply theorem 27.10. Thus,
Multiplying opposite pieces of the same chord, we get:
PA×PB = PD×PC ⇒ 2×2 = 1×PC ⇒ PC = 4
9. So diameter of the bangle = CD = PD + PC = 1 + 4 = 5 cm
10. So radius of the bangle = ⁵⁄₂ = 2.5 cm

Solved example 27.19
Find the value of 'x' in each of the three cases in fig.27.54 below:

Fig.27.54

Solution:
1. Consider fig(a). We can multiply opposite pieces: 16×6 = x×12 ⇒ 96 = 12x ⇒ x = 8 
2. Consider fig(b). We can multiply opposite pieces: 20×6 = x×8 ⇒ 120 = 8x ⇒ x = 15
3. Consider fig(c). We can multiply opposite pieces: But length of one piece is not given
■ We are given two clues:
• The chord with a total length of 13 cm, is a diameter. Because it is passing through the centre 'O'
• This diameter intersects the other chord in a perpendicular direction. So the other chord is bisected. 
    ♦ That means, the length of the other piece is also 'x'. See Theorem 17.1.
So we can write: 9×4 = x×x ⇒ 36 = x² ⇒ x = 6

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In the next section, we will see another application of theorem 27.10.

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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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Chapter 27 - Additional Examples

Additional Example 1
In the figure (a) below, A, B, C and D are points on the circle. 

Compute the angles of the quadrilateral ABCD, and the angles between it's diagonals
Solution:
1. Consider ΔBPC. We get ∠BPC = [180-(30+50)] =[180-80] = 100^o. (∵ sum of interior angles of a triangle is 180^o) This is marked in fig(b)
2. ∠BPC and ∠BPA form a linear pair. So ∠BPA = 180 - ∠BPC = 180 -100 = 80^o.

3. ∠BPC and ∠APD are opposite angles, and are hence equal. So we get ∠APD = ∠BPC = 100^o.

4 Similarly, ∠BPA and ∠DPC are opposite angles, and are hence equal. So we get ∠DPC = ∠BPA = 80^o.

5. Consider arc BC in fig(c). It subtends ∠BAC (= 35^o) on the alternate arc. 

• The same arc subtends ∠BDC on the alternate arc. So ∠BDC = ∠BAC = 35^o.

6. Consider ΔPCD. We get ∠PCD = [180-(80+35)] =[180-115] = 65^o. (∵ sum of interior angles of a triangle is 180^o)

7. Consider arc CD in fig(c). It subtends ∠CBD (= 30^o) on the alternate arc. 

• The same arc subtends ∠CAD on the alternate arc. So ∠CAD = ∠CBD = 30^o.

8. Consider ΔPAD. We get ∠ADP = [180-(100+30)] =[180-130] = 50^o. (∵ sum of interior angles of a triangle is 180^o)

■ Thus we get all the angles of the quadrilateral ABCD. Since it is a cyclic quadrilateral, we can do a check:

(i) Sum of opposite angles ∠BAD and ∠BCD = 35 + 30 + 50 + 65 = 180^o.          

(ii) Sum of opposite angles ∠ABC and ∠ADC = 65 + 30 + 50 + 35 = 180^o.          

Additional Example 2

In the figure (a) below, AB is a diameter of the circle. 

CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E. Calculate ∠AEB

Solution:

1.Draw OC and OD as shown in fig(b). Consider the ΔOCD

• OC = OD (∵ radii of the same circle)

• Given that chord CD is equal to the radius.

• So ΔOCD is an equilateral triangle. All it's interior angles are equal to 60^o.

2. Draw AD as shown in fig(c)

3. Consider the arc CD. It has a central angle ∠COD = 60^o.

4. This same arc CD subtends ∠CAD on the opposite arc. 

• So ∠CAD = ¹⁄₂×60 = 30^o.  

5. Now consider ∠ADB. AB is a diameter and D is a point on the semi-circle. So ADB = 90^o.

6. Consider ΔAED. The ∠ADB that we considered above is an exterior angle of ΔAED

• Exterior angle of a triangle = sum of remote interior angles

• So ∠ADB = ∠EAD + ∠AED ⇒ 90 = 30 + ∠AED ⇒ ∠AED = 90 - 30 = 60^o.

7. But ∠AED and ∠AEB are the same angles

• Thus the required ∠AEB = 60^o

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