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 17.6 - Circles and chords - solved examples

In the previous section we completed the discussion on some of the properties of circles. We also saw some solved examples. In this section, we will see some solved examples that are related to the topics that we discussed so far in this chapter in general.

Solved example 17.13

Find the radius of the circle shown in fig.17.30(a). AB is 24 cm long. CD is the perpendicular bisector of AB, and it is 8 cm long.

Fig.17.30

Solution:

1. CD is the perpendicular bisector of the chord AB. So if it is extended inwards, it will meet the centre ‘O’ of the circle. (See theorem 17.1)

2. Let the radius OB = r cm

3. Consider ⊿OBC. Applying Pythagoras theorem, we get: r²-12² = x²

4. Also we have OD = r = 8+x

5. Substituting this value of r in (3), we get: (8+x)²-12² = x² ⇒ 64 + 16x + x² -12² = x²

⇒ 64 + 16x - 144 = 0 ⇒ 16x = 144 – 64 = 80. So x = 80/16 = 5

6. Substituting this value of x in (4), we get: r = 8 + 5 = 13 cm

Solved example 17.14

In the fig.17.31(a), AB is the diameter of the circle. PQ is parallel to AB. 

Fig.17.31

If AB = 50 cm and PQ = 14 cm, find BQ

Solution:

1. AB is the diameter. So radius r = 25 cm. PQ is a chord. It is shown in magenta colour in fig.b.

2. Perpendicular distance of PQ from O can be obtained by rearranging Eq.17.1: p = √[r²-(^c⁄₂)²]

3. So we get: OD = p = √[25²-7²] = √[625-49] = √576 = 24

4. QR is drawn perpendicular to AB.

5. QR = OD = p = 24 cm (∵ AB and PQ are parallel)

6. Consider ⊿QRB.

• QR = p [from (5)]

• RB = OB – OR = 25 – 7 = 18

7. Applying Pythagoras theorem we get: BQ = √[QR² + RB²] = √[24² + 18²] = √[576 + 324] = √[900] = 30

Solved example 17.15

In the fig.17.32(a), OE = 6 cm and ED = 15 cm. Find the radius of the circle

Fig.17.32

Solution:

1. BD is a chord, and OE is a perpendicular drawn from the centre to this chord

2. So, the perpendicular OE will split the chord into two equal parts BE and DE. (See theorem 17.1)

3. Thus BE = DE = 15 cm (∵ it is given that BE = 15 cm)

4. Consider ⊿OBE. Applying Pythagoras theorem we get: OB = √[OE² + BE²] = √[6² + 15²] 

= √[36 + 225] = √[261] = 16.2 cm

Solved example 17.16

From a circular disc of radius 17 cm, a portion in the shape of a trapezium is cutoff. Parallel sides are 30 cm and 16 cm. Both the parallel sides are on the same side of the centre of the circle. Find the area of the trapezium.

Solution:

1. Fig.17.33(a) shows the given data. 

Fig.17.33

2. In fig.b, the trapezium is named as ABCD. The parallel sides AB and CD become two chords of the circle. Radius r of the circle = 17 cm

3. OP and OQ are the perpendicular distances of the two chords from the centre. They can be obtained rearranging Eq.17.1:

p = √[r²-(^c⁄₂)²]

4. So we get OP = √[17²-15²] = √[289-225] = √64 = 8

5. OQ = √[17²-8²] = √[289-64] = √225 = 15

6. Area of a trapezium = ¹⁄₂ × (top side + bottom side) × height

7. Here height of trapezium = PQ = OQ – OP = 15 – 8 = 7

8. So area = ¹⁄₂ × (30 + 16) × 7 = 161 sq.cm

Solved example 17.17
Three points A, B and C are situated at equal distances from each other, along the periphery of a circle. The radius of this circle is 20 m. What is the distance between the points?
Solution:
1. Given that the distances between the points are equal. So they form the vertices of an equilateral triangle.
2. Also, the three points are on the periphery of a circle. So the circle is the circumcircle.
3. We have seen the relation between the side of an equilateral triangle and the radius of it's circumcircle. The relation is S = r√3. Where S is the side and r is the circumradius
4. So we get: Distance between the points = S = 20√3 = 20 × 1.732 = 34.64 m
5. The actual construction is shown in fig.17.34 below:

Fig.17.34

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This completes our present discussion on circles. Links to some solved examples in the form of video presentations are given below:
Solved example 17.18      Solved example 17.19       Solved example 17.20
In the next section, we will see 'Parallel lines'.

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Chapter 17.4 - Length of Chords

In the previous section we saw properties related to 'equal chords'. In this section, we will see  'length of chords'.

Length of chords

Consider fig.17.18 below. AB is a chord. OP is the perpendicular from the centre. OA is the radius r.

Fig.17.18

• OAP is a right angled triangle. Applying Pythagoras theorem, we get: AP = √[OA² – OP²]. But AP is 'half the chord'. So this is an easy method to get the length of any chord. We will write it in the form of a theorem:

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Theorem 17.5:

■ Half of chord is the square root of:

    • the difference of two quantities:

        ♦ Square of the radius

        ♦ square of the perpendicular distance of the chord from the centre

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The steps can be combined to form a single equation:
Eq.17.1:

^c⁄₂ = √[r²- p²]

Where c is the length of chord, r is the radius of circle and p is the perpendicular distance of the chord from the centre

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Once we get 'the half', we will get 'the full', just by multiplying it with '2'. Let us see an example: 
A circle has a radius of 4 cm. One of it’s chords is situated at a perpendicular distance of 3 cm from the centre. What is the length of the chord?

Solution:

Applying theorem 17.5, we have:

Half of the chord is the square root of:

  • the difference of two quantities:

      ♦ square of radius = 4² = 16

      ♦ square of the perpendicular distance of the chord from the centre = 3² = 9

• So difference = 16-9 = 7

• So half of the chord is √7.

• Thus full chord = 2√7

Another method: We can avoid the detailed steps by using Eq.17.1:
• Using Eq.17.1, we have: ^c⁄₂ = √[r²- p²]
• Given r = 4 cm and p = 3 cm

• ^c⁄₂ = √[4²- 3²] = √[16- 9] = √7

• So c = 2√7 cm

Another type of problem:

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:

1. Fig.17.19(a) shows the piece of bangle. The distance between it’s ends A and B is 4 cm. It’s height is 1 cm.

Fig.17.19

2. Fig.b shows the full bangle. AB is taken as a chord. The height 1 cm is measured in a direction perpendicular to the chord AB. 
3. OP is the perpendicular from the centre to AB. OP is extended further until it meets the circle at Q. So OQ is the radius. O and B are joined by the line OB. 
4. OPB forms a right triangle. 
• It’s hypotenuse OB = r. 
• One leg OP = (r -1) cm. 
• Other leg PB = 2 cm. 
5. Applying Pythagoras theorem we get:

• r² – (r-1)² = 2² ⇒ r² – (r² - 2r + 1) = 4 ⇒ (r² – r² + 2r - 1) = 4 ⇒ 2r - 1 = 4 ⇒ 2r = 5
• Thus we get r = 2.5 cm

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Now we will see some solved examples

Solved example 17.6

In a circle, a chord 1 cm away from the centre is 6 cm long. What is the length of the chord which is 2 cm away from the centre?

Solution:

A rough sketch is shown in fig.17.20. Note that it is only a rough sketch. Actual construction using the given measurements is not required to solve the problem.

Fig.17.20

1. Given: length of chord c = 6 cm. So ^c⁄₂ = 3 cm

2. Also given: perpendicular distance p = 1 cm. These details relate to the lower ⊿OAP in fig.17.20

3. Using Eq.17.1, we have: ^c⁄₂ = √[r²- p²]

4. r is the unknown. So we must bring it to the left side.

• ^c⁄₂ = √[r²- p²] ⇒ (^c⁄₂)² = [r²- p²] (squaring both sides)

• ⇒ r² = [(^c⁄₂)² + p²] ⇒ r = √[(^c⁄₂)² + p²]

5. Thus we get r = √[3² + 1²] = √[9 + 1] = √10

• Now we use this r for the second part of the problem. This is related to the ⊿OCQ

6. We have ^c⁄₂ = √[r²- p²] = √[(√10)²- 2²] = √[10 - 4] = √6

7. So c = 2√6

Solved example 17.7

In a circle of radius 5 cm, two parallel chords of lengths 6 and 8 cm are drawn on either side of the diameter. What is the distance between them? If parallel chords of these lengths are drawn on the same side of a diameter, what would be the distance between them?

Solution:

Part I: Chords on either side of diameter

A rough sketch is shown in fig.17.21(a) below: 

Fig.17.21

Note that it is only a rough sketch. Actual construction using the given measurements is not required to solve the problem.

1. Consider the chord of 8 cm length. It is named as AB in the fig.17.21(a)

2. Given: length of chord c = 8 cm. So ^c⁄₂ = 4 cm

3. Also given: radius r = 5 cm. These details relate to ⊿OAP in fig.a

4. We have ^c⁄₂ = √[r²- p²]. Here p is the unknown. So we must bring it to the left side. 
5. ^c⁄₂ = √[r²- p²] ⇒ (^c⁄₂)² = [r²- p²] (squaring both sides)

⇒ p² = [ r² - (^c⁄₂)²] ⇒ p = √[ r² - (^c⁄₂)²]

6. Thus we get OP = p = √[5² - 4²] = √[25 - 16] = √9 = 3

7. Consider the chord of 6 cm length. It is named as CD in the fig.a

8. Given: length of chord c = 6 cm. So ^c⁄₂ = 3 cm

9. Also given: radius r = 5 cm. These details relate to ⊿OCQ in fig.b

10. We have p = √[ r² - (^c⁄₂)²]

11. Thus we get OQ = p = √[5² - 3²] = √[25 - 9] = √16 = 4

12. So the distance between the two chords = PQ = 3 + 4 = 7 cm

Part II: Chords on the same side of the diameter

• A rough sketch is shown in fig.c. Note that it is only a rough sketch. Actual construction using the given measurements is not required to solve the problem.

1. AB is the chord of 8 cm. We know that equal chords will be at equal distances from the centre. 
2. We have seen in part I of this problem that, the chord of 8 cm length is at a distance of 3 cm from the centre. 
3. So all chords of 8 cm length (in this particular circle of 5 cm radius) will be at 3 cm from the centre, what ever be the position. 
4. So we do not need to do the calculations. In fig.c, OP = 3 cm

5. Similarly, from part I, OQ = 4 cm. Thus, the distance between the two chords = PQ = OQ – OP = 4 – 3 = 1 cm

Solved example 17.8

The bottom side of the quadrilateral in the fig.17.22(a) is a diameter of the circle.

Fig.17.22

The top side is parallel to the bottom side. Calculate the area of the quadrilateral.

Solution:

1. The bottom side of the quadrilateral is a diameter. The top side is a chord.

2. Given that the top side is parallel to the bottom side. So it is a trapezium

3. Area of a trapezium = ¹⁄₂ × (top side + bottom side) × height

4. As the top and bottom sides are parallel, and the bottom side is the diameter, the 'height of the trapezium' is the perpendicular distance of the chord from the centre

5. In fig.b, the chord is named as AB, and the perpendicular distance is named as OP

6. We have ^c⁄₂ = √[r²- p²]. p is the unknown. So we must bring it to the left side. 
7. ^c⁄₂ = √[r²- p²] ⇒ (^c⁄₂)² = [r²- p²] (squaring both sides)

⇒ p² = [ r² - (^c⁄₂)²] ⇒ p = √[ r² - (^c⁄₂)²]

8. Thus we get OP = p = √[2.5² – 1.5²] = √[6.25 – 2.25] = √4 = 2

9. So from (3) we get: Area of a trapezium = ¹⁄₂ × (3 + 5) × 2= 8 sq.cm

Solved example 17.9

In a circle, two parallel chords of lengths 4 and 6 cm are 5 cm apart. What is the radius of the circle?

Solution:

A rough sketch is shown in fig.17.23. Note that it is only a rough sketch. Actual construction using the given measurements is not necessary to solve the problem.

Fig.17.23

1. In the fig., chord AB has a length of 6 cm, and chord CD has a length of 4 cm

2. The perpendicular distances are OP and OQ. We are given the total distance PQ as 5 cm

3. Let the distance OP be 'x' cm. Then OQ = (5-x) cm

4. In the upper ⊿OAP we have r² = 3² + x² ⇒ r² = 9 + x²

5. In the lower ⊿OCQ we have r² = 2² + (5-x)² ⇒ r² = 4 + 25 – 10x + x² ⇒ r² = 29 – 10x + x²

6. Both (4) and (5) gives r². So we can equate them:

7. 9 + x² = 29 – 10x + x² ⇒ 10x = 20. So x = 2

8. Substituting this value of x in (4) we get:
9. r² = 9 + 2² ⇒ r² = 9 + 4 ⇒ r² = 13. So r = √13

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In the next section, we will see 'Circles through given points'.

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