1.4.1 Circular Measure

In this topic we will learn to:

• understand the definition of a radian, and use the relationship between radians and degrees
• use the formulae $s = r\theta$ and $A = \frac{1}{2}r^{2}\theta$ in solving problems concerning the arc length and sector area of a circle

$$\textbf{\textcolor{gray}{Radian}}$$A radian is an angle whose corresponding arc in a circle is equal to the radius of the circle.

One radian is $\approx 57.2958$ degrees.

To change an angle from radians to degrees, we use the formula,
$$\theta = \textmd{angle in radians} \times \frac{180^{\circ}}{\pi}$$To change an angle from degrees to radians, we use the formula,
$$\theta = \textmd{angle in degrees} \times \frac{\pi}{180^{\circ}}$$$$\textbf{\textcolor{gray}{Arc Length}}$$An arc is a portion of the circumference of the circle. The length of that portion is called the arc length. Arc length is represented by the symbol, $s$. To calculate the arc length, we use the formula,
$$s = r\theta$$Where $r$ is the radius of the circle and $\theta$ is the angle of the sector.
$$\textbf{\textcolor{gray}{Sector Area of a Circle}}$$A sector of a circle is a pie shaped portion of a circle, consisting of an arc and its two radii. To calculate the sector area of circle, we use the formula,
$$A = \frac{1}{2}r^{2}\theta$$Note: The formulae for arc length and sector area of a circle take $\theta$ in radians NOT in degrees.

Let’s look at some past paper questions on this topic.

1. A sector of a circle of radius $r$ cm has an area of $A$ cm$^{2}$. Express the perimeter of the sector in terms of $r$ and $A$. (9709/11/M/J/19 number 3)

Perimeter is the distance around the shape,
$$P = r + \textcolor{#2192ff}{s} + r$$Using the formula for arc length, substitute $s$ with $r\theta$,
$$P = r + \textcolor{#2192ff}{r\theta} + r$$$$P = 2r + r\theta$$Use the formula for sector area of a circle, to get rid of $\theta$,
$$A = \frac{1}{2}r^{2}\theta$$Make $\theta$ the subject of the formula,
$$\theta = \frac{2A}{r^{2}}$$Substitute $\theta$,
$$P = 2r + r\textcolor{#2192ff}{\theta}$$$$P = 2r + r\left(\textcolor{#2192ff}{\frac{2A}{r^{2}}}\right)$$Simplify,
$$P = 2r + \frac{2A}{r}$$Therefore, the final answer is,
$$P = 2r + \frac{2A}{r}$$2. The diagram shows a sector $ABC$ of a circle with center $A$ and radius $r$. The line $BD$ is perpendicular to $AC$. Angle $CAB$ is $\theta$ radians. (9709/11/M/J/22 number 5)

(a) Given that $\theta = \frac{1}{6}\pi$, find the exact area of $BCD$ in terms of $r$.

If you look at the diagram, you will notice that the Area of $BCD$ can be written as,
$$\textmd{Area of }BCD = \textmd{Area of Sector }ABC -\textmd{ Area of triangle }ABD$$Use the formula for sector area of a circle and area of a triangle,
$$\textmd{Area of }BCD = \frac{1}{2}r^{2}\theta - \frac{1}{2} \times BD \times AD$$Substitute in the value of $\theta$,
$$\textmd{Area of }BCD = \frac{1}{2}r^{2}\left(\frac{1}{6}\pi\right) - \frac{1}{2} \times BD \times AD$$Simplify,
$$\textmd{Area of }BCD = \frac{1}{12}\pi r^{2}\theta - \frac{1}{2} \times BD \times AD$$Let’s evaluate $BD$ using Pythagoras,
$$\sin{\theta} = \frac{BD}{r}$$$$BD = r\sin{\theta}$$Since $\theta = \frac{1}{6}$, then,
$$BD = r\sin\left(\frac{1}{6}\pi\right)$$$$BD = \textcolor{#2192ff}{\frac{1}{2}r}$$Let’s evaluate $AD$ using Pythagoras,
$$\cos{\theta} = \frac{AD}{r}$$$$AD = r\cos{\theta}$$$$AD = r\cos\left(\frac{1}{6}\pi\right)$$$$AD = \textcolor{#0f0}{\frac{\sqrt{3}}{2}r}$$Let’s substitute $BD$ and $AD$,
$$\textmd{Area of }BCD = \frac{1}{12}\pi r^{2}\theta - \frac{1}{2} \times \textcolor{#2192ff}{BD} \times \textcolor{#0f0}{AD}$$$$\textmd{Area of }BCD = \frac{1}{12}\pi r^{2}\theta - \frac{1}{2} \times \textcolor{#2192ff}{\frac{1}{2}r} \times \textcolor{#0f0}{\frac{\sqrt{3}}{2}r}$$Simplify,
$$\textmd{Area of }BCD = \frac{1}{12}\pi r^{2}\theta - \frac{\sqrt{3}}{8}r^{2}$$Therefore, the final answer is,
$$\textmd{Area of }BCD = \frac{1}{12}\pi r^{2}\theta - \frac{\sqrt{3}}{8}r^{2}$$(b) Given instead that the length of $BD$ is $\frac{\sqrt{3}}{2}r$, find the exact perimeter of $BCD$ in terms of $r$.

Perimeter is the distance around the shape,
$$P = \textcolor{#2192ff}{BD} + CD + arcBC$$Substitute $BD$ with $\frac{\sqrt{3}}{2}r$,
$$P = \textcolor{#2192ff}{\frac{\sqrt{3}}{2}r} + CD + arcBC$$From the diagram, we can tell that $CD = AC - AD$,
$$P = \frac{\sqrt{3}}{2}r + \textcolor{#0f0}{(AC - AD)} + arcBC$$From part $(a)$ we know that $AD = r\cos{\theta}$ and $AC$ is the radius,
$$P = \frac{\sqrt{3}}{2}r + \textcolor{#0f0}{(r - r\cos{\theta})} + arcBC$$Using the formula for arc length, substitute $arcBC$ with $r\theta$,
$$P = \frac{\sqrt{3}}{2}r + (r - r\cos{\theta}) + \textcolor{#2192ff}{r\theta}$$We can use Pythagoras to evaluate $\theta$,
$$\sin{\theta} = \frac{BD}{AB}$$Note: The right-angled triangle allows us to use SOHCATOA.
$$\sin{\theta} = \frac{\frac{\sqrt{3}}{2}r}{r}$$$$\sin{\theta} = \frac{\sqrt{3}}{2}$$$$\theta = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$$$$\theta = \frac{1}{3}\pi$$Substitute $\theta$,
$$P = \frac{\sqrt{3}}{2}r + (r - r\cos{\textcolor{#2192ff}{\theta}}) + r\textcolor{#2192ff}{\theta}$$$$P = \frac{\sqrt{3}}{2}r + \left(r - r\cos\left(\textcolor{#2192ff}{\frac{1}{3}\pi}\right)\right) + r\left(\textcolor{#2192ff}{\frac{1}{3}\pi}\right)$$Simplify,
$$P = \frac{\sqrt{3}}{2}r + \left(\textcolor{#0f0}{r - \frac{1}{2}r}\right) + \frac{1}{3}\pi r$$$$P = \frac{\sqrt{3}}{2}r + \textcolor{#0f0}{\frac{1}{2}r} + \frac{1}{3}\pi r$$Therefore, the final answer is,
$$P = \frac{\sqrt{3}}{2}r + \frac{1}{2}r + \frac{1}{3}\pi r$$3. In the diagram, $CXD$ is a semicircle of radius $7$ cm with center $A$ and diameter $CD$. The straight line $YABX$ is perpendicular to $CD$, and the arc $CYD$ is part of a circle with center $B$ and radius $8$ cm. Find the total area of the region enclosed by the two arcs. (9709/12/F/M/19 number 3)

From the diagram, we can tell that the total area can be written as,
$$\textmd{Total Area } = \textmd{Area of semicircle }CXD + \textmd{Area of segment }CYD$$Let’s start by finding the Area of semicircle $CXD$,
$$\textmd{Area of semicircle }CXD = \frac{1}{2}\pi r^{2}$$$CXD$ has a radius of $7$ cm,
$$\textmd{Area of semicircle }CXD = \frac{1}{2}\pi (7)^{2}$$Simplify,
$$\textmd{Area of semicircle }CXD = \frac{49}{2}\pi$$Let’s find the Area of segment $CYD$,
$$\textmd{Area of segment } CYD = \textmd{Area of sector }CYD - \textmd{Area of triangle }BCD$$Use formulae for sector area of a circle and area of a triangle,
$$\textmd{Area of segment } CYD = \frac{1}{2}r^{2}\theta - \frac{1}{2} \times CD \times AB$$Substitute in $r$ and $CD$,
$$\textmd{Area of segment } CYD = \frac{1}{2} \times 8^{2} \times \theta - \frac{1}{2} \times 14 \times AB$$Simplify,
$$\textmd{Area of segment } CYD = 32\theta - 7\times AB$$Let’s find $AB$ using Pythagoras,
$$\textcolor{#2192ff}{a}^{2} + \textcolor{#0f0}{b}^{2} = \textcolor{red}{c}^{2}$$$$(\textcolor{#2192ff}{AB})^{2} + (\textcolor{#0f0}{AC})^{2} = (\textcolor{red}{BC})^{2}$$$$(AB)^{2} + 7^{2} = 8^{2}$$$$(AB)^{2} = 8^{2} - 7^{2}$$$$(AB)^{2} = 15$$$$AB = \sqrt{15}$$Substitute $AB$ and simplify,
$$\textmd{Area of segment } CYD = 32\theta - 7\times \textcolor{#2192ff}{AB}$$$$\textmd{Area of segment } CYD = 32\theta - 7\times \textcolor{#2192ff}{\sqrt{15}}$$$$\textmd{Area of segment } CYD = 32\theta - 7\sqrt{15}$$Let’s find $\theta$,
$$\theta = 2ABC$$$$\sin(ABC) = \frac{AC}{BC}$$$$\sin(ABC) = \frac{7}{8}$$$$ABC = \sin^{-1}\left(\frac{7}{8}\right)$$$$\theta = 2\sin^{-1}\left(\frac{7}{8}\right)$$Substitute $\theta$ and simplify,
$$\textmd{Area of segment } CYD = 32\textcolor{#2192ff}{\theta} - 7\sqrt{15}$$$$\textmd{Area of segment } CYD = 32\left(\textcolor{#2192ff}{2\sin^{-1}\left(\frac{7}{8}\right)}\right) - 7\sqrt{15}$$$$\textmd{Area of segment } CYD = 64\sin^{-1}\left(\frac{7}{8}\right) - 7\sqrt{15}$$Therefore, the total area is,
$$\textmd{Total Area } = \textcolor{#2192ff}{\textmd{Area of semicircle }CXD} + \textcolor{#0f0}{\textmd{Area of segment }CYD}$$$$\textmd{Total Area }= \textcolor{#2192ff}{\frac{49}{2}\pi} + \textcolor{#0f0}{64\sin^{-1}\left(\frac{7}{8}\right) - 7\sqrt{15}}$$$$\textmd{Total Area }= 118.0$$Note: Remember that $\theta$ is in radians, so evaluate any trig functions in radians.

Therefore, the final answer is,
$$\textmd{Total Area }= 118.0$$