Trigonometric functions and circular measure - the Australian ...

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{32} • Trigonometric functions and circular measureExampleIn a circle of radius 36 cm, find the area of a sector subtending an angle of 30 ◦ at thecentre.SolutionO30 º36 cmWith r = 36 and θ = 30 ◦ = π , we have6A = 1 2 × 362 × π 6 = 108π cm2 .As mentioned above, in problems such as these it is best to leave your answer in termsof π unless otherwise stated.The area A s of a segment of a circle is easily found by taking the difference of two areas.OθrIn a circle of radius r , consider a segment that subtends an angle θ at the centre. We canfind the area of the segment by subtracting the area of the triangle (using 1 2ab sinθ) fromthe area of the sector. ThusA s = 1 2 r 2 θ − 1 2 r 2 sinθ = 1 2 r 2 (θ − sinθ).
A guide for teachers – Years 11 and 12 • {33}ExampleFind the area of the segment shown.18 º4OSolutionWith r = 4 and θ = 18 ◦ = 18π180 = π , we have10A s = 1 ( π2 × 42 ×10 10)− sin π ≈ 0.041.Exercise 15Around a circle of radius r , draw an inner and outer hexagon as shown in the diagram.By considering the perimeters of the two hexagons, show that 3 ≤ π ≤ 2 3.Summary of arc, sector and segment formulasLength of arcArea of sectorArea of segmentl = r θA = 1 2 r 2 θA = 1 2 r 2 (θ − sinθ)
Page 1 and 2: 1Supporting Australian Mathematics
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Page 5 and 6: A guide for teachers - Years 11 and
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