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A segment of a circle is the region between a chord and the arc subtended by that chord. Consider chord AB subtended by central angle \(\theta\) in a circle with radius \(r\).Derive a formula using \(r\) and \(\theta\) for the area of the segment subtended by \(\theta.\)

Short Answer

Expert verified
\[ \text{Area of the segment} = \frac{1}{2} r^2 (\theta - \sin(\theta)) \]

Step by step solution

01

Understand the Circle Segment

A circle segment is the area between a chord and the corresponding arc. To find the area of the segment, first calculate the area of the sector formed by the central angle \( \theta \) and radius \( r \).
02

Area of the Sector

The area of a sector with central angle \( \theta \) radians and radius \( r \) is given by: \[ \text{Area of the sector} = \frac{1}{2} r^2 \theta \]
03

Area of the Triangle

The area of the triangular region formed by the chord AB and the radii to the endpoints of the chord can be calculated using the formula: \[ \text{Area of the triangle} = \frac{1}{2} r^2 \sin(\theta) \]
04

Subtract the Triangle Area

To find the area of the segment, subtract the area of the triangular region from the area of the sector: \[ \text{Area of the segment} = \frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin(\theta) \]
05

Simplify the Formula

Factor out \( \frac{1}{2} r^2 \) from the expression to get the final formula: \[ \text{Area of the segment} = \frac{1}{2} r^2 (\theta - \sin(\theta)) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circle Segment
A circle segment is the portion of a circle outlined by a chord and the arc it intercepts. In easier terms, imagine cutting a pizza slice and then cutting off the crust. The remaining piece without the crust is a circle segment.
To calculate the area of a circle segment, there are several steps involved. These steps help to break down the problem and make it manageable. We need to work with both the sector area and the triangle area to find our solution.
Sector Area
The sector of a circle looks like a pizza slice. It’s the region enclosed by two radii (the slices) and the arc (crust). The sector area depends on the radius of the circle and the central angle.
For a circle with radius \(r\) and central angle \(\theta\) measured in radians, the formula for the sector area is:
\[ \text{Area of the sector} = \frac{1}{2} r^2 \theta \]
This formula is derived from the fact that the total area of a circle is \(\pi r^2\) and the sector represents a fraction \(\frac{\theta}{2 \pi}\) of the entire circle based on the angle \(\theta\).
Triangle Area
To find the area of the segment, we first need to calculate the area of the triangle formed by the chord AB and the two radii that connect to the chord's endpoints.
The formula for the area of such a triangle, with base formed by the chord and vertex angle \(\theta\), is:
\[ \text{Area of the triangle} = \frac{1}{2} r^2 \sin(\theta) \]
This formula uses the sine of the central angle \(\theta\), and it comes from basic trigonometry.
Central Angle in Radians
In these formulas, the angle \(\theta\) is expressed in radians rather than degrees. Radians offer a natural way to work within the context of circles because they relate directly to the radius.
A full circle is \(2\pi\) radians, which is approximately 6.28 radians. Here's a quick conversion:
  • 90 degrees = \(\frac{\pi}{2}\) radians
  • 180 degrees = \(\pi\) radians
  • 360 degrees = \(2\pi\) radians
In calculating the area, it is vital to use radians to ensure the formulas work correctly. For example, if \(\theta\) were expressed in degrees, the formulas for the sector and triangle areas would not hold.

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Most popular questions from this chapter

Determine the exact roots for each trigonometric equation or statement in the specified domain. a) \(2 \cos \theta-\sqrt{3}=0,0 \leq \theta < 2 \pi\) b) \(\csc \theta\) is undefined, \(0^{\circ} \leq \theta < 360^{\circ}\) c) \(5-\tan ^{2} \theta=4,-180^{\circ} \leq \theta \leq 360^{\circ}\) d) \(\sec \theta+\sqrt{2}=0,-\pi \leq \theta \leq \frac{3 \pi}{2}\)

Alice Through the Looking Glass by Lewis Carroll introduced strange new worlds where time ran backwards. Your challenge is to imagine a unit circle in which a positive rotation is defined to be clockwise. Assume the coordinate system remains as we know it. a) Draw a unit circle in which positive angles are measured clockwise from \((0,1) .\) Label where \(R\left(\frac{\pi}{6}\right), R\left(\frac{5 \pi}{6}\right), R\left(\frac{7 \pi}{6}\right)\) and \(\mathrm{R}\left(\frac{11 \pi}{6}\right)\) are on your new unit circle. b) What are the coordinates for the new \(\mathrm{R}\left(\frac{\pi}{6}\right)\) and \(\mathrm{R}\left(\frac{5 \pi}{6}\right) ?\) c) How do angles in this new system relate to conventional angles in standard position? d) How does your new system of angle measure relate to bearings in navigation? Explain.

Determine the exact measure of all angles that satisfy the following. Draw a diagram for each. a) \(\sin \theta=-\frac{1}{2}\) in the domain \(\mathbf{0} \leq \boldsymbol{\theta}<2 \pi\) b) \(\cot \theta=1\) in the domain \(-\pi \leq \theta<2 \pi\) c) \(\sec \theta=2\) in the domain \(-180^{\circ} \leq \theta<90^{\circ}\) d) \((\cos \theta)^{2}=1\) in the domain \(-360^{\circ} \leq \theta<360^{\circ}\)

Sketch each angle in standard position. In which quadrant does each angle terminate? a) 1 b) \(-225^{\circ}\) c) \(\frac{17 \pi}{6}\) d) \(650^{\circ}\) e) \(-\frac{2 \pi}{3}\) f) \(-42^{\circ}\)

Nora is required to solve the following trigonometric equation. \(9 \sin ^{2} \theta+12 \sin \theta+4=0, \theta \in\left[0^{\circ}, 360^{\circ}\right)\) Nora did the work shown below. Examine her work carefully. Identify any errors. Rewrite the solution, making any changes necessary for it to be correct. \(9 \sin ^{2} \theta+12 \sin \theta+4=0\) \((3 \sin \theta+2)^{2}=0\) $$ 3 \sin \theta+2=0 $$ Therefore. \(\sin \theta=-\frac{2}{3}\) Use a colculator. \(\sin ^{-1}\left(-\frac{2}{3}\right)=-41.8103149\) So, the reference ongle is 41.8 , to the neorest tenth of a degree Sine is negotive in quodrants II ond III. The solution in quadront II is \(180^{\circ}-41.8^{\circ}=138.2\) The solution in quadrant III is \(180^{\circ}+41.8=221.8\) Therefore, \(\theta=138.2^{\circ}\) ond \(\theta=221.8\), to the neorest tenth of a degree.

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