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Geometry and Trigonometry - The trigonometric circle – Arcs and Sectors

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Radian Measure: A radian is the angle subtended at the center of a circle by an arc equal in length to the radius. For any circle, the circumference is 2πr2\pi r, meaning a full revolution is 2π2\pi radians. Therefore, 180∘=π180^\circ = \pi radians.

Diagram showing a circle where arc length equals radius, defining 1 radian.
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Arc Length and Sector Area: For a circle with radius rr and central angle θ\theta (in radians), the arc length ss is proportional to the angle (s=rθs = r\theta). The sector area is the fraction of the total area πr2\pi r^2 corresponding to the angle θ2π\frac{\theta}{2\pi}, simplifying to A=12r2θA = \frac{1}{2}r^2\theta.

A sector of a circle highlighting the central angle theta, the arc length s, and the shaded area.
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The Unit Circle: A circle centered at the origin (0,0)(0,0) with radius r=1r = 1. The coordinates of any point PP on the circle are (cos⁡θ,sin⁡θ)(\cos \theta, \sin \theta), where θ\theta is the angle measured from the positive xx-axis. This relates geometry to the Pythagorean identity cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1.

Unit circle on a coordinate plane showing a point P at coordinates (cos theta, sin theta).
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Segments of a Circle: A segment is the region bounded by a chord and its corresponding arc. Its area is calculated by subtracting the area of the triangle formed by the two radii and the chord (Atriangle=12r2sin⁡θA_{triangle} = \frac{1}{2}r^2 \sin \theta) from the area of the whole sector.

Diagram of a circle segment showing the chord and the resulting area between the chord and the arc.

📐Formulae

θradians=θdegrees×π180\theta_{radians} = \theta_{degrees} \times \frac{\pi}{180}

s=rθs = r\theta

Asector=12r2θA_{sector} = \frac{1}{2}r^2\theta

Asegment=12r2(θ−sin⁡θ)A_{segment} = \frac{1}{2}r^2(\theta - \sin \theta) (where θ\theta is in radians)

x2+y2=1x^2 + y^2 = 1 (Equation of the unit circle)

💡Examples

Problem 1:

A circle has a radius of 66 cm. Find the length of an arc that subtends an angle of 120∘120^\circ at the center. Give your answer in terms of π\pi.

Solution:

  1. Convert the angle from degrees to radians: θ=120∘×π180=2π3 radians\theta = 120^\circ \times \frac{\pi}{180} = \frac{2\pi}{3} \text{ radians}
  2. Use the arc length formula s=rθs = r\theta: s=6×2π3s = 6 \times \frac{2\pi}{3} s=4π cms = 4\pi \text{ cm}

Explanation:

First, we convert the given degree measure into radians because the formula s=rθs = r\theta requires θ\theta to be in radians. Then, we substitute the radius and the angle into the formula.

Problem 2:

In a circle with radius r=4r = 4 cm, a sector has an area of 88 cm2\text{cm}^2. Find the angle θ\theta of the sector in radians and the perimeter of the sector.

Solution:

  1. Use the area formula A=12r2θA = \frac{1}{2}r^2\theta to find θ\theta: 8=12(4)2θ8 = \frac{1}{2}(4)^2\theta 8=8θ8 = 8\theta θ=1 radian\theta = 1 \text{ radian}
  2. Find the arc length ss: s=rθ=4×1=4 cms = r\theta = 4 \times 1 = 4 \text{ cm}
  3. Calculate the perimeter PP (sum of two radii and the arc length): P=r+r+s=4+4+4=12 cmP = r + r + s = 4 + 4 + 4 = 12 \text{ cm}

Explanation:

We rearrange the sector area formula to solve for the unknown angle θ\theta. Once θ\theta is found, we calculate the arc length and add it to two lengths of the radius to find the total perimeter of the sector.

Problem 3:

Calculate the area of a segment of a circle with radius 1010 cm and a central angle of π3\frac{\pi}{3} radians.

Solution:

  1. Use the segment area formula: Asegment=12r2(θ−sin⁡θ)A_{segment} = \frac{1}{2}r^2(\theta - \sin \theta)
  2. Substitute r=10r = 10 and θ=π3\theta = \frac{\pi}{3}: Asegment=12(10)2(π3−sin⁡(π3))A_{segment} = \frac{1}{2}(10)^2\left(\frac{\pi}{3} - \sin\left(\frac{\pi}{3}\right)\right)
  3. Simplify the expression: Asegment=50(π3−32)A_{segment} = 50\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right) Asegment=50π3−253≈9.06 cm2A_{segment} = \frac{50\pi}{3} - 25\sqrt{3} \approx 9.06 \text{ cm}^2

Explanation:

The area of the segment is found by subtracting the area of the triangle 12r2sin⁡θ\frac{1}{2}r^2\sin \theta from the area of the sector 12r2θ\frac{1}{2}r^2\theta. Since sin⁡(π/3)=3/2\sin(\pi/3) = \sqrt{3}/2, we substitute the exact values to find the area.

Problem 4:

A windshield wiper of length 5050 cm rotates through an angle of 2π3\frac{2\pi}{3} radians. Calculate the area of the windshield cleaned by the blade, assuming the blade starts at a distance of 1010 cm from the center of rotation.

Diagram showing the area cleaned by a wiper as a sector with an inner hole.

Solution:

  1. The region cleaned is an annulus sector (a large sector minus a small sector).
  2. Outer radius R=50+10=60R = 50 + 10 = 60 cm.
  3. Inner radius r=10r = 10 cm.
  4. Angle θ=2π3\theta = \frac{2\pi}{3}.
  5. Area A=12R2θ−12r2θ=12θ(R2−r2)A = \frac{1}{2}R^2\theta - \frac{1}{2}r^2\theta = \frac{1}{2}\theta(R^2 - r^2).
  6. A=12×2π3×(602−102)=π3×(3600−100)=3500π3≈3665.19A = \frac{1}{2} \times \frac{2\pi}{3} \times (60^2 - 10^2) = \frac{\pi}{3} \times (3600 - 100) = \frac{3500\pi}{3} \approx 3665.19 cm2\text{cm}^2.

Explanation:

We treat the cleaned area as the difference between two sectors sharing the same central angle but having different radii.

Problem 5:

A chord ABAB divides a circle of radius 88 cm into two segments. If the length of the chord is 88 cm, find the area of the minor segment.

Circle with an equilateral triangle formed by two radii and a chord of length 8.

Solution:

  1. Since radius r=8r=8 and chord c=8c=8, the triangle OABOAB (where OO is the center) is equilateral.
  2. Therefore, the central angle θ=60∘=π3\theta = 60^\circ = \frac{\pi}{3} radians.
  3. Area of sector OAB=12r2θ=12×82×π3=32π3≈33.51OAB = \frac{1}{2}r^2\theta = \frac{1}{2} \times 8^2 \times \frac{\pi}{3} = \frac{32\pi}{3} \approx 33.51 cm2\text{cm}^2.
  4. Area of triangle OAB=12r2sin⁡θ=12×64×sin⁡(π3)=32×32=163≈27.71OAB = \frac{1}{2}r^2 \sin \theta = \frac{1}{2} \times 64 \times \sin\left(\frac{\pi}{3}\right) = 32 \times \frac{\sqrt{3}}{2} = 16\sqrt{3} \approx 27.71 cm2\text{cm}^2.
  5. Area of segment =33.51−27.71=5.80= 33.51 - 27.71 = 5.80 cm2\text{cm}^2.

Explanation:

By identifying the triangle as equilateral, we find the central angle. Subtracting the triangle's area from the sector's area yields the segment's area.