# back to Trigonometry lessons

### Minor arc and Central Angle In the figure above, point O is the center of a circle. Points A and B lie on the circle, so OA and OB are two radiuse. Since the vertex of the angle AOB is the center of the circle, so, the angle AOB is the central angle. The center angle AOB opposite the minor arc AB. If the length of the arc AB is equal to the length of the radius of the circle, then the central angle opposite the arc AB is 1 rad. The angle a is the angle AOB.

### Definition of 1 Rad of the Central Angle In the figure above, if the length of the arc AB is equal to the radius of the circle, then measure of the central angle AOB is 1 rad.

### Perimeter and Central Angle Look the figure above, the point O is the center of the circle with radius r. If OA counterclockwise rotate to a full circle, then the central angle a is 360o. The perimeter of the circle is 2pi r.

### Conversion between Degree and Radian In the figure above, when the ray OA counteclockwise rotate a full circle, the central angle formed is 360o, which is 2pi.

when the ray OA counteclockwise rotate half circle, the central angle AOG is (1/2) × 360o = 180o, which is (1/2) × 2pi = pi.

when the ray OA counteclockwise rotate quater circle, the central angle AOD is (1/4) × 360o, which is (1/4) × 2pi = pi/2.

when the ray OA counteclockwise rotate three-quater circle, the central angle AOJ is (3/4) × 360o = 270o, which is (3/4) × 2pi = (3/2)pi.

when the ray OA counteclockwise rotate to the OB position, the central angle AOB is (1/3) × 90o = 30o, which is (1/3) × pi/2 = pi/6.

when the ray OA counteclockwise rotate to the OC position, the central angle AOC is (2/3) × 90o = 60o, which is (2/3) × pi/2 = pi/3.

when the ray OA counteclockwise rotate to the OE position, the central angle AOE is 90o + 30o = 120o, which is pi/2 + pi/6 = 2pi/3.

when the ray OA counteclockwise rotate to the OF position, the central angle AOF is 90o + 2 × 30o = 150o, which is pi/2 + 2 × pi/6 = 5pi/6.

when the ray OA counteclockwise rotate to the OH position, the central angle AOH is 180o + 30o = 210o, which is pi + pi/6 = 7pi/6.

when the ray OA counteclockwise rotate to the OI position, the central angle AOI is 180o + 2 × 30o = 240o, which is pi + 2 × pi/6 = 4pi/3.

when the ray OA counteclockwise rotate to the OK position, the central angle AOK is 270o + 30o = 300o, which is 3pi/2 + pi/6 = 5pi/3.

when the ray OA counteclockwise rotate to the OL position, the central angle AOL is 270o + 2 × 30o = 330o, which is 3pi/2 + 2 × pi/6 = 11pi/6.

# Transformation between Degree and Radian Example 1
Solution
multiply 135o by pi/180o.
135o × pi/180o = = 3pi/4 [note: both numerator and denominator divide by 45]
Example 2
convert 5pi/4 to the degree measure
Solution
5pi/4 × 180o/pi = 225o

# Sector   ### Length of an arc In the figure above, l is the length of the arc AB, which opposite the central angle AOB and r is the radius of the circle. The length of the arc AB is the product of the central angle and radius r. That is, l = |a|r

note: Since the length of the arc AB is always positive, so, there is the absolute value of the central angle in the formula of the length of the arc.

note: Use the formula of the length of the arc, the central angle uses the radian as unit.

Example 3 In the figure above, the central angle AOB is 30o, the arc AB is 2 feet. What is the radius r of the circle?

Solution

Step 1: Covert the central angle to the radian measure. Angle AOB = 30o × pi/180o = pi/6

Step 2: Use the formula of the length of the arc, l = |a|r, so, r = l /|a| = 2 ÷ pi/6 = 2 × 6/pi = 12/pi = 3.82. Therefore, the radius of the circle is 3.82 feet.

Example 4

The perimeter of a sector is 32 inches. Find its central angle a, which make the sector has a maximum area. Solution

Given: the perimeter of the sector AOB is 32 inches. That is, 2r + l = 32, in which l is the length of the arc AB. So, the length of the arc AB is, l = 32 - 2r

Let A is the area of the sector AOB, then A = (1/2) × r × l, in which l is the length of the arc AB.

So, A = (1/2) r (32 - 2r) = 16 r - r2. Therefore, the area of a sector is a function of the radius of the circle.

A = - (r2 - 16r)
= - (r2 - 16r + 82 - 82)
= - (r2 - 16r + 82) + 64
= - (r - 8)2 + 64

From the equation above, r is always a positive number. The first term of the area of the sector AOB is a negative number and second term is a positive number. When r = 8, the first term is zero. At this condition, A is maximum and its value is 64 square inches.

when r = 8, A is maximum and A = 64 square inches.
when r = 8,
the central angle a = l/r
= (32 - 2r)/r
= (32 - 2 × 8)/8
= (32 - 16)/8 = 16/8
Therefore, when the central angle a is 2 rad, the sector AOB has the maximum area.

# The Length of the Chord

Find the relation between the length of the chord and the central angle that is opposite the chord. In the figure above, point O is the center of the circle with radius r. Points A and B lie on the circle. Since OA = OB = r, so triangle OAB is an isosceles triangle. Draw line segment OC perpendicular to AB, then OC bisects the central angle AOB which is the angle a. So the angle AOC = the angle BOC = (1/2) a. And also, AC bisects the chord AB. So, AC = BC = (1/2) AB. Since OC is perpendicular to AB, so triangle AOC and triangle BOC are right triangles.

In right triangle AOC,
sin(a/2) = AC/r
AC = r sin(a/2)
2 AC = 2r sin(a/2)
AB = 2r sin(a/2)

So, the length of the chord AB is the product of the diameter of the circle and sine half of the central angle.

# The area of a sector Let A be the area of the sector AOB, then
A = (1/2) r l = (1/2) a r2

In the formula above, r is the radius of the circle, l is the length of the arc AB and a is the central angle that opposite the arc AB.

Example 5 In the figure above, the length of the chord AB is 6 inches and the central angle that the chord opposite is 4 rad. Find the area of the sector AOB.

Solution

Use the formula we get from the length of the chord section
AB = 2 r sin(a/2)
given: AB = 6 and a = 4 rad
6 = 2 r sin(4/2)
6 = 2 r sin(2)
both side of the equation divide by 2
3 = r sin(2)
r = 3/sin(2)
the area of the sector AOB is,
A = (1/2) a r2
= (1/2) × 4 × [3/sin(2)]2
= 2 × 9/sin2(2)
= 18/sin2(2)
convert 2 rad to degree measure
2 × 180o/pi = 114.65o
sin(114.65)o = 0.91
0.912 = 0.83
18/0.83 = 21.69

Therefore, the area of the sector AOB is 21.69 square inches.

Example 6

The perimeter of the sector AOB is 12 inches. The central angle a is 1 rad. What is the area of the sector? Solution

Let P be the perimeter of the sector AOB, then
P = 2 r + l, in which, l is the length of the arc AB
given P = 12, then
12 = 2 r + l, then l = 12 - 2 r
by the formula, l = a r
so, l = 1 × r = r
so, r = 12 - 2 r
move the - 2r to the left side of the equation
r + 2 r = 12
3 r = 12
r = 4
let A be the area of the sector AOB, then
A = (1/2) r l = (1/2) × 4 × 4 = 8

Therefore, the area of the sector AOB is 8 square inches.

Example 7

The central angle a is 135o and the radius of the circle is 12 inches, what is the shadow area? Solution

the central angle a is
a = 135o × pi/180o = 3pi/4
the length of the arc AB is
l = a r = 3pi/4 × 12 = 9pi
the area of the sector AOB is
Area of sector AOB = (1/2) r l
= (1/2) × 12 × 9pi
= 54pi
= 169.56 square inches
Finf the area of the triangle AOB

Draw a line from point B perpendicular to the extended line AO, which marked as h that is the height of the triangle AOB to the base AO.

h = r sin(180o - 135o)
= r sin(45)o
= 12 × (square root of 2)/2
= 6 × square root of 2
= 8.48
The area of the triangle AOB is
Area of the triangle AOB = (1/2) AO × h
= (1/2) × 12 × 8.48
= 50.88 square inches
The area of the shadow = the area of the sector AOB - the area of the triangle AOB
= 169.56 - 50.88
= 116.68 square inches.

Therefore, the area of the shadow is 116.68 square inches.