One radian is the angle at the centre of a sector whose arc length equals its radius. So in a circle of radius 1, an arc of length 1 subtends exactly 1 radian.

Define an arc and a sector of a circle.

An arc is a part of the circumference , the curved edge on its own. A sector is the region enclosed by two radii and an arc, so it is the slice rather than just its crust.

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Circular Measure (Cambridge (CIE) IGCSE Additional Maths): Flashcards

Exam code: 0606

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Define a radian.

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Cards in this collection (14)

Define a radian.
One radian is the angle at the centre of a sector whose arc length equals its radius.
So in a circle of radius 1, an arc of length 1 subtends exactly 1 radian.
How do you convert between degrees and radians?
Everything follows from $π$ radians $= 180 °$ .
To go from radians to degrees multiply by $\frac{180}{π}$ , and to go from degrees to radians multiply by $\frac{π}{180}$ .
Complete these common conversions, in degrees:
$\frac{π}{2} =______$
$\frac{π}{3} =______$
$\frac{π}{4} =______$
$\frac{π}{6} =______$
The completed conversions are:
$\frac{π}{2} = 90 °$
$\frac{π}{3} = 60 °$
$\frac{π}{4} = 45 °$
$\frac{π}{6} = 180 ° \div 6 = 30 °$
Each is $180 °$ divided by the number underneath, which is worth knowing well enough to build others from.
Convert $43.8 °$ to radians, giving an exact answer.
Divide by $180 °$ and multiply by $π$ :
$\frac{43.8}{180} \times π = \frac{73}{300} π$
Leaving it as a fraction of $π$ keeps the answer exact, and it is about $0.764$ as a decimal.
True or False?
You may leave the units off an angle in radians, but never off an angle in degrees.
True.
An angle written as a multiple of $π$ is understood to be in radians, so $\frac{π}{6}$ needs no label, though rad may be written.
A bare number could be either, so a degree measurement always needs its $^{\circ}$ symbol.
Why is it easier to work in radians when a question already involves $π$ ?
Because the angles are themselves multiples of $π$ , so the $π$ terms usually cancel during the working.
Converting to degrees would replace an exact multiple of $π$ with a decimal, which then has to be rounded.
Define an arc and a sector of a circle.
An arc is a part of the circumference, the curved edge on its own.
A sector is the region enclosed by two radii and an arc, so it is the slice rather than just its crust.
What makes an arc or sector minor rather than major?
The angle at the centre: less than $180 °$ gives the minor one, more than $180 °$ gives the major one.
The minor sector is a single slice of the pizza, and the major sector is everything left after that slice is taken.
Complete the arc length formula, where $θ$ is in radians:
$l =______$
The completed formula is:
$l = r θ$
It is as simple as this only because radians are defined so that an arc of length $r$ subtends 1 radian.
Where does $l = r θ$ come from?
An arc is a fraction of the whole circumference, and in radians that fraction is $\frac{θ}{2 π}$ .
So $l = \frac{θ}{2 π} \times 2 π r$ , and the $2 π$ cancels to leave $r θ$ .
A slice of angle $\frac{π}{6}$ is cut from a circular pizza of radius 12 cm. Find the perimeter of the piece left behind.
The remaining angle is $2 π - \frac{π}{6} = \frac{11 π}{6}$ , so the major arc is $12 \times \frac{11 π}{6} = 22 π$ .
The perimeter also includes the two straight cut edges, each a radius, giving $22 π + 24$ cm.
Complete the sector area formula, where $θ$ is in radians:
$A =______$
The completed formula is:
$A = \frac{1}{2} r^{2} θ$
It comes from taking the fraction $\frac{θ}{2 π}$ of the whole area $π r^{2}$ and cancelling.
A sector of radius 6 cm has area 30 cm². Find the angle at the centre, in radians.
Substitute into $A = \frac{1}{2} r^{2} θ$ :
$30 = \frac{1}{2} \times 36 \times θ$
So $18 θ = 30$ , giving $θ = \frac{5}{3}$ radians.
True or False?
The formulas $l = r θ$ and $A = \frac{1}{2} r^{2} θ$ work whether $θ$ is in degrees or radians.
False.
Both are derived using $2 π$ for a full turn, so they hold only when $θ$ is in radians.
With degrees you need the longer forms, $l = \frac{θ}{360} \times 2 π r$ and $A = \frac{θ}{360} \times π r^{2}$ .

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