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29 Applications of Radian Measure

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29 Applications of Radian Measure - 29 Applications of Radian Measure This section focuses on the use of Radians to measure angles and the appli- cations of this, including areas and arc-lengths. By the end of this section, you should have the following skills: • An understanding of definition of a Radian. • Convert angles between degrees and radians. • An understanding of the definition of a sector and segment. • Calculate the arc length and area of a circle sector and apply this to real life examples. 29.1 Radians Radians are often used instead of degrees when measuring angles. In degrees a complete revolution of a circle is 360◦, however in radians it is 2π. If an arc of a circle is drawn such that the radius is the same length as the arc, the angle created is 1 Radian (as shown below). 1 29.2 Converting Between Degrees and Radians • Degrees to Radians: π Radians = × Degrees 180 • Radians to Degrees: 180 Degrees = × Radians π Example 1 1. Convert the following angles from degrees to radians. (a) 140◦ (b) 20◦ (c) 270◦ 2. Convert the following angles from radians to degrees. 2 (a) 3 π (b) 3π 16 (c) 9 π Solution: 1. (a) π 7 140◦ = × 140 = π Rad 180 9 (b) π 1 20◦ = × 20 = π Rad 180 9 (c) π 3 270◦ = × 270 = π Rad 180 2 2. (a) 2 180 2 180 × 2 π = × π = = 120◦ 3 π 3 3 2 (b) 180 3π = × 3π = 180 × 3 = 540◦ π (c) 16 180 16 180 × 16 π = × π = = 320◦ 9 π 9 9 29.3 Arc Length of a Circle The length of an arc is just a portion of the circumference, which we know to be 2πr, where r is the radius of the circle. This give us the equation for arc length of: l = θr where r is the radius of the circle and θ is the angle of the arc, in radians. Example 2 Find the length of the arc of a segment cut out of a circle with a radius of 4 cm and an angle of 0:7 radians. Solution: l = θr = 0:7 × 4 = 2:8 cm Exercise 1 (a) An arc AB with center O has a radius of 7 and an angle of 60◦. Convert the angle into radians and calculate the length of the arc 3 AB. 2π (b) A piece of pizza has a radius of 9 cm and an angle of 3 . Find the perimeter of the piece of pizza. Solutions to exercise 1 (a) Firstly, converting 60◦ into radians, we have: π π 60◦ = × 60 = Rad 180 3 We can now use the equation below to calcualte the arc length: π 7 l = θr = × 7 = π = 7:330 (3 d.p.) 3 3 (b) We must first calculate the arc length of the pizza slice: 2π l = θr = × 9 = 6π 3 For the perimeter we must add the two straight sides of the pizza slice, which both have a length of r, to the arc length we have cal- culated above: L = 6π + 2r = 6π + 18 = 36:850 (3 d.p.) 29.4 Sector Area The formula for the area of a sector of a circle is much simpler when using radians. For a sector of a circle with radius r and angle θ in radians, we have the following area: 1 A = r2θ 2 4 π Example 3 Find the area of a sector with angle θ = 6 and radius r = 3. Solution: Substituting these values into the equation above, we have: 1 1 π 3π A = r2θ = × 32 × = 2 2 6 4 29.5 Segment Area The diagram below shows a major and minor segment. The area of a segment of a circle, such as the shaded area of the sketch above can be calculated using radians. For a circle with radius r the area of a segment with an angle of θ is: 1 A = r2(θ − sin θ) 2 Example 4 In the diagram below AB is the diameter of a circle with a radius r, with an angle α in radians. Given that the area of the triangle 4AOC is 4 times the area of the shaded segment. Use this and the equation for the area of a segment get an equation in terms of α, and 4 show that sin α = 5 α. 5 Solution: The area of the segment is given by: 1 r2(α − sin α) 2 The area of the triangle 4AOC is given by: 1 1 r2 sin(π − α) = r2 sin α 2 2 as sin(180◦ − θ◦) = sin θ so sin(π − θ) = sin θ. Now as the area of the triangle is 4 times the shaded area, we have: 1 1 r2 sin α = 4 × r2(α − sin α) 2 2 sin α = 4(α − sin α) 4 sin α = α 5 Exercise 2 (a) For the circles below, find the area of the sector shaded in green to 6 3 d.p., given that α, β and λ have the values. 4 5 1 α = π; β = π; λ = π 3 12 4 (b) The diagram below is a swimming pool. Find the perimeter and area 4 of the swimming pool floor, given that α = 9 π. Solutions to exercise 2 (a) (a) The shaded area is a sector of a circle with a radius of 7 cm, 7 4 and with an angle of α = 3 π, and so the area is given by: 1 A = r2θ 2 1 4 = × 72 × π 2 3 = 102:625 cm2 (3 d.p.) (b) The shaded area is a sector of a circle with a radius of 3 cm, 5 and with an angle of β = 12 π, and so the area is given by: 1 A = r2θ 2 1 5 = × 32 × π 2 12 = 1:875π = 5:890 cm2 (3 d.p.) (c) The shaded area is a sector of a circle with a radius of 9 cm, and 7 with an angle of 2π −λ = 4 π (this is because we are calculating the area of the major sector but λ relates to the angle in the minor sector), and so the area is given by: 1 A = r2θ 2 1 7 = × 92 × π 2 4 = 70:875π = 222:660 cm2 (3 d.p.) (b) The perimeter of the swimming pool: AB and DC both have a length of 12 m, and we can calculate the length of the arcs AD and BC once we know the length of the 8 diagonals. Using Pythagoras' Theorem, BD2 = AD2 + AB2 p BD = 82 + 122 p = 208 p = 4 13 p Hence AE = 2 13, and so we have: AD = BC = AE × α p 4 = 2 13 × π 9 = 10:069 (3 d.p.) Hence the perimeter of the pool is: P = AB + BC + CD + DA = (2 × 12) + (2 × 10:069) = 44:138 m (3 d.p.) The area of the swimming pool: The area is given by the rectangle ABCD plus two segments of equal area at AD and BC. We can calculate these areas then add them to get the total area of the pool: • The area of the rectangle ABCD: 12 × 8 = 96 m2 9 • The area of the segment at AB: 1 = AE2(α − sin α) 2 1 p 2 4 4 = × 2 13 × π − sin π 2 9 9 4 4 = 26 π − sin π 9 9 = 10:698 m2 (3 d.p.) Hence the area of the pool floor is given by: A = 96 + 2 × 10:698 = 117:396 m2 10 29.6 Videos Arc length and area of a circle sector Finding the length of a circular arc and the area of the corresponding circle sector. 11.
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