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 21.10 - Solved examples on Circles, Arcs and Sectors

In the previous section we completed the discussion on Area of sectors. In this section we will see some solved examples related to this chapter in general.

Solved example 21.25
Perimeter of two circles are in the ratio 2:3
(i) What is the ratio of their radii?
(ii) What is the ratio of their areas?
Solution:
1. Let the radii of the two circles be r₁ and r₂. Then the perimeters are 2πr₁ and 2πr₂.
2. Given that 2πr₁ : 2πr₂ = 2:3
3. Dividing both sides by 2π we get r₁ : r₂ = 2:3 [This is the answer for part(i)]
4. Ratio of areas = πr₁² : πr₂² 
5. Dividing both sides of the ratio by π, we get: 
Ratio of areas = r₁² : r₂²  
6. From (3) we have: ^r₁⁄_r2 = ²⁄₃ ⇒ r₁ = ^2r₂⁄₃ . Substituting this value of r₁ in (5), we get:
7. Ratio of areas = (^2r₂⁄₃)² : r₂²
8. Dividing both sides of the ratio by r₂² we get:
Ratio of areas = ⁴⁄₉ :1 ⇒ Ratio of areas = 4:9 [This is the answer for part(ii)]

Solved example 21.26
A wheel of radius 20 cm rolls along. How much would it move ahead after 10 rotations?
Solution:
Consider the circle in fig.21.43 below. The circle represents a wheel

Fig.21.43

1. Position (1) shows the position when the wheel is just about to roll to the right. At that moment, the point of contact of the wheel with the ground is marked as 'A'. 
2. When the wheel rolls, point A moves along the circumference of the wheel. When A touches the ground again, one rotation is completed. This is shown as position (2)
3. So the distance travelled after one rotation is equal to the perimeter of the wheel
4. So the distance travelled after 10 rotations = 10 × perimeter = 10 × 2πr = 20πr
5. In this problem, r is given as 20 cm. So the distance travelled = 20 × 3.14 × 20 = 1256 cm

Solved example 21.27
Two semicircular pieces are cut off from a rectangular sheet as shown in the fig.21.44 below.

Fig.21.44

 Find the area of the remaining part
Solution:
1. Total area = area of the rectangle = 50 × 20 = 1000 cm².
2. Consider any one semicircular part. It reaches from the top edge to bottom edge of the rectangle.
3. So the diameter of the semicircle = height of the rectangle = 20 cm
4. So radius = 10 cm
5. Area of the semicircular part = half the area of a circle with 10 cm radius
= ¹⁄₂ × πr² = ¹⁄₂ × π × 10² = 50π cm²
6. There are two semicircular parts. Their total area = 2 × 50π = 100π cm².
7. So area of the remaining part = 1000 - 100π = 1000 - (100×3.14) = 1000 - 314 = 686 cm²

Solved example 21.28
The length of an arc of a circle is 4 cm. What is the length of an arc with the same central angle, in a circle of double the radius?
Solution:
1. We are given the length of an arc. No other details are given. So let us assume the required details:
Let the radius be r, and let the central angle be x
2. So length of arc will be equal to (^πr⁄₁₈₀ × x) cm
3. But the length of arc is given as 4 cm. So we can write:
(^πr⁄₁₈₀ × x) = 4 
4. Now we are given another arc. It has the same central angle. But radius = 2r. We want the length of this new arc
5. We can write: length of new arc = (^π(2r)⁄₁₈₀ × x)
⇒ length of new arc = 2 × (^πr⁄₁₈₀ × x)
6. But from (3), the quantity inside brackets is 4 cm. So we get:
7. length of new arc = 2 × 4 = 8 cm

Solved example 21.29
In the fig.21.45 below, the sector OAB has a radius of 6 cm, and central angle 60^o.

Fig.21.45

Find the area of the blue region
Solution:
1. If a sector has central angle 60^o, we can separate out an equilateral triangle from it.

2. Area after the removal of equilateral triangle from a sector of radius 'r' and central angle 60^o

= [(^π⁄₆) -  (^√3⁄₄)]r²  cm². Details here.

3. In our problem, r = 6 cm. So we get:

Area = [(^π⁄₆) -  (^√3⁄₄)]6² =  [6π - 9√3] cm²

Solved example 21.30
The area of a circular table is 31400 cm². What is it's radius? And it's perimeter?
Solution:
1. Area of a circle = πr² = 3.14 × r² = 31400
⇒ r² = ³¹⁴⁰⁰⁄_3.14 = 10000 ⇒ r = √10000 = 100 cm
2. Perimeter = 2πr = 2 × 3.14 × 100 = 628 cm 

Solved example 21.31
In fig.21.46 below, OPQR is a sector of a circle with centre O and radius 8 cm.

Fig.21.46

Find the area of the yellow region.
Solution:
1. The central angle of the sector OPQR is given as 90^o. So this sector will cover one fourth of the area of the full circle. We can write:
Area of sector OPQR = ¹⁄₄ × πr² = ¹⁄₄ × π × 8² = 16π cm².
2. Now, OPR is a triangle. It is a right triangle. We have it's base and height. So area of ΔOPR
= ¹⁄₂ × b × h = ¹⁄₂ × 8 × 8 = 32 cm².
3. So area of the yellow region = (16π - 32) = 16(π - 2) cm².


Solved example 21.32
The length of the pendulum of a clock is 12 cm. It makes an angle of 45^o in one swing. What is the distance travelled by the tip of the pendulum in on swing?
Solution:
1. Fig.21.47 below shows one swing of the pendulum

Fig.21.47

2. In one swing, the pendulum travels from P to Q. It is an arc of radius 12 cm, and central angle 45^o.
3. So we want the length of arc PQ. We can use theorem 21.1
According to the theorem, For every 1^o central angle, the length of arc will be (^πr⁄₁₈₀) cm 
4. So for 45^o, the length of arc will be (^π×12⁄₁₈₀) × 45 =  (^π×12⁄₄) = 3π cm

Solved example 21.33
In the fig.21.48 below, 'O' is the centre of the circle, and OPQR is a rectangle.

Fig.21.48

OP = 8 cm and OR = 15 cm. Find the area of the yellow region
Solution:
1. Consider a diagonal OQ of the rectangle. It's length can be obtained as follows:
OQ² = OP² + PQ²
⇒ OQ² = (8)² + (15)² = 6•4 + 225 = 289 (∵ PQ = OR = 15 cm)

⇒ OQ = √289 = 17
2. But OQ is the radius of the circle. So we have all the details required to find the area of the yellow region:
• It's radius = 17 cm
• It's central angle = 90^o
3. The yellow region is one fourth of the total area of the circle. So it's area
=  ¹⁄₄ × πr² = ¹⁄₄ × π × 17² = 72.25π cm².

Solved example 21.34
In the fig.21.49 below, both sectors OAB and OPQ have the same centre O, and the same central angle 45^o.

Fig.21.49

The sum of the radii is 18 cm. Area of the shaded portion ABQP is 18π. Find the radius of both the sectors.
Solution:
1. If we subtract the area of the inner sector OPQ from the area of the outer sector, we will get the area of the shaded portion ABQP
2. Area of the inner sector OPQ = (^πr2⁄₃₆₀) × x = (^π×OP2⁄₃₆₀) × 45 = (^π⁄₈)×OP² cm²
3. Area of the outer sector OAB = (^πr2⁄₃₆₀) × x = (^π×OA2⁄₃₆₀) × 45 = (^π⁄₈)×OA² cm². 
4. So area of ABPQ = [(^π⁄₈)×OA² - (^π⁄₈)×OP² ] = (^π⁄₈)×[OA² - OP² ] cm².
5. But area of ABPQ is given as 18π. So we can equate the two:
 (^π⁄₈)×[OA² - OP² ] = 18π.
⇒[OA² - OP² ] = 18 × 8 = 144  
⇒(OA-OP)(OA+OP) = 144 [∵ (a² - b²) = (a+b)(a-b)]
6. But (OA+OP) is given as 18 cm. So we can write:
(OA-OP) × 18 = 144 ⇒ (OA-OP) = 8 cm. Now we have two equations:
7. OA + OP = 18
8. OA - OP = 8 
9. From (8) we get: OA = 8 + OP
10. Substituting this value of OA in (7), we get:
8 + OP + OP = 18 ⇒ 8 + 2OP = 18 ⇒ 2OP = 10 ⇒ OP = 5 cm
11. Substituting this value of OP on (8) we get: OA = 8 + 5 = 13 cm

Solved example 21.35
Find the area of the yellow portion in fig.21.50 below

Fig.21.50

Solution:
1. Area of the green sector = (^πr2⁄₃₆₀) × x = (^π×62⁄₃₆₀) × 60 = 6π cm² 
2. Total area = πr² = π × 6²  = 36π cm².
3. So area of the yellow portion = 36π - 6π = 30π cm².
Another method:
1. Central angle of the yellow portion = 360 -60 = 300^o.
2. So area = (^πr2⁄₃₆₀) × x = (^π×62⁄₃₆₀) × 300 = 30π cm² 

Solved example 21.36
Given a circular metal disc of perimeter 30 cm. A regular hexagon of maximum possible size is to be cut out from the disc. What will be the perimeter of the hexagon?
Solution:
Consider fig.21.51 below

Fig.21.51

1. A regular hexagon can be divided into 6 equilateral triangles by drawing it's three diagonals.
2. Each diagonal will be a diameter of the circle. So our first step is to find the diameter of the circle
3. Given that perimeter = 30 cm. That means 2πr = 30. So diameter = 2r = ³⁰⁄_π .
4. Each side of the equilateral triangle = half of diameter = radius = ¹⁄₂ × ³⁰⁄_π = ¹⁵⁄_π .
5. So side of the regular hexagon = ¹⁵⁄_π cm.
6. Thus perimeter of the regular hexagon = 6 × ¹⁵⁄_π = ⁹⁰⁄_π cm

Solved example 21.37
In the fig.21.52 below, O is the centre of the circle, and AB is the diameter 

Fig.21.52

The length of an arc with central angle 72^o is 4π cm.
(i) Find the diameter of the circle.
(ii) Find the area of the green portion
Solution:
1. Given that: length of an arc with central angle 72^o is 4π cm.
2. From this, we can find the radius: We can use theorem 21.1. According to the theorem, For every 1^o central angle, the length of arc will be (^πr⁄₁₈₀) cm 
3. So for 72^o, the length of arc will be (^π×r⁄₁₈₀) × 72 =  (^π×r⁄₅) × 2 cm
4. But this length is given as 4π cm. So we can equate the two:
(^π×r⁄₅) × 2 cm = 4π. ⇒ r = 10 cm 
5. So diameter of the circle = 20 cm. This is the answer of part (i)
6. AB is the diameter. So central angle of the yellow portion = 180^o.
7. So central angle of the green portion = [360 -(180 + 72)] = 108^o.
8. So area of the green portion = (^πr2⁄₃₆₀) × x = (^π×102⁄₃₆₀) × 108 = 30π cm². This is the answer of part (ii)

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In the next section we will see Real numbers.

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