# Circles, Sectors and Trigonometry Problems with Solutions and Answers

Sectors and Circles trigonometry problems are presented along with their detailed solutions.

## Coordinates and Angles in Standard Position

When solving the problems below, we will make use of the definition of the trigonometric functions of angle θ in terms of the coordinates of point P(x , y) that is on the terminal side of an angle θ.
cos(θ) = x / R     tan(θ) = y / x     sec(θ) = R / x
sin(θ) = y / R     cot(θ) = x / y     CSC(θ) = R / y
R = √(x2 + y2)
We will also use the relationship between the arc length S, the radius R and the central angle θ of a sector: S = R θ     (θ in RADIANS)
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## Problems on Circles and Arcs

### Problem 1

If the coordinates of point A, in the circle below, are (8, 0) and arc s has a length of 20 units, find the coordinates of point P in the diagram below.
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Solution
Radius R of the given circle is equal to 8. We first use the formula for the arc s in terms of the radius R and the angle θ to find θ in radians.
s = R θ
2 0 = 8 θ
θ = 20 / 8 = 2.5 radians
We now use the definitions for sine and cosine to find the coordinates x and y of P.
cos(θ) = x / R     and     sin(θ) = y / R
cos(θ) = x / R     gives     x = R cos(θ) = 8 cos(2.5) = -6.40
sin(θ) = y / R     gives     y = R sin(θ) = 8 sin(2.5) = 4.79
Point P has the coordinates (-6.40 , 4.79).

### Problem 2

In a rectangular coordinate system, a circle with its center at the origin passes through the point(4√2 , 5√2).
What is the length of the arc S?
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Solution
We use the definiton for tangent to find angle θ.
tan(θ) = y / x = 5√2 / 4√2 = 5 / 4
θ = arctan(5 / 4)
Use fromula for s in terms of θ and the radius R.
S = R θ = √ (4√2)2 + (5√2)2 arctan(5 / 4) ≈ 8.11

### Problem 3

The length of the minutes hand in a clock is 4.5 cm. Find the length of the arc traced by the end of the minutes hand between 11:10 pm and 11:50 pm.
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Solution
From 11:10 to 11:50, there
11:50 - 11:10 = 40 minutes
One complete rotation of the minutes hand corresponds to 60 minutes and to an angle of 360°. We therefore can write that there are
360 ° / 60 minutes = 6 ° / minutes
If θ is the central angle corresponding to the arc traced by the minutes hand, then it is given by
θ = (6 °/ minute) × 40 minutes = 240 °
The arc S traced is then given by
S = R θ , (θ in radians).
S = R θ = 4.5 cm 240 × π / 180 = 6π

### Problem 4

The points P(a , b), Q(5 , 0) and M(c , -1) are located on the circle with the center O(0 , 0) and radius R of 5 units as shown below. Calculate the:
a) The coordinates (a,b) of point P.
b) The arc length between Q and P in counterclockwise direction.
c) The x-coordinate c of point M.
d) Angle θ.
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Solution
a) Point P is the terminal side of an angle in standard position of size 35°. Hence
a = R cos (35°) = 5 cos (35°)     and     b = R sin(35°) = 5 sin (35°)
b) The central angle QOP is known, therefore the length of the arc PQ is given by
arc PQ = r×35 × π / 180 = 5×35 × π / 180 ≈ 3.05 units
c) Point C is on the circle, therefore the distance from the center of the circle (0,0) to point C is equal to the radius.
(c - 0)2 + (-1 - 0)2 = 5
Square both sides of the above equation and solve for c.
(c - 0)2 + (-1 - 0)2 = 25
c = ~+mn~ √ 24 = ~+mn~ 2√ 6
Point M is in quadrant III and therefore c is negative. Hence point M has the coordinates
M(- 2√ 6, -1)
d) Point M is on the terminal side of angle θ+35°. Hence using the definition of the tangent of an angle in standard position in terms of the coordinates of a point on the terminal side, we have
tan(θ + 35 °) = 1 / 2√ 6
Point M is in quadarnt III, therefore θ + 35 ° is greater than 180° and therefore
θ + 35 ° = 180 + arctan(1 / 2√ 6)
θ = 180 + arctan(1 / 2√ 6) - 35 ° = 156.54°

### Problem 5

Circle of radius R = 4 units and center O is shown below.
a) Find the length of the arc between D and P.
b) Find the coordinates (a,b) of P.
c) Find the length of segment PD,.
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Solution
a) The central angle DOP is known, therefore length of the arc DP is given by (do not forget to convert degrees to radians)
arc PQ = R 130 π / 180 = 4 × 130 π / 180 ≈ 9.08 units
b) Point P is on the terminal side of an angle in standard position. Therefore the coordinates are given by:
a = R cos(130°) = 4 cos(130°)     and     b = R sin(130°) = 4 sin(130°)
c) Use the distance formula between two points P and D to find the length of the line segment PD.
PD = √ (a - 4)2 + (b - 0)2
= √ (4 cos(130°) - 4)2 + (4 sin(130°) - 0)2 ≈ 7.25

### Problem 6

In the circle of center O and radius R = 3 shown below, the length of the arc S between A and B is 12.5 units. Find the coordinates (a , b) of point B.

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Solution
We first need to find the size θ of the central angle AOB.
S = R θ
θ = S / R = 12.5 / 3 radians
We now calculate the coordinates a and b of point B.
a = R cosθ = 3 cos(12.5 / 3)
b = R cosθ = 3 sin(12.5 / 3)

### Problem 7

Point P has coordinates (4.1 , b) and is located on the circle, of center O and radius R = 5, in quadrant I. Find the length of arc S.
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Solution
We first need to calculate the size θ of the central angle DOP using the x coordinate of point P and the radius.
4.1 = 5 cos(θ)
cos(θ) = 4.1 / 5
θ = arccos(4.1 / 5)
We calculate the arc S.
S = R θ = 5 arccos(4.1 / 5)