Radian Measure: past paper questions & worked solutions

For a sector, $A=\frac12 r^2\theta$. Here $A=15$ and $\theta=1.25=\frac54$, so $15=\frac12 r^2\cdot\frac54$.

$r^2=\dfrac{30}{5/4}=24$, so $r=\sqrt{24}=2\sqrt6$ because the radius is positive.

Arc length is $s=r\theta=(2\sqrt6)\left(\frac54\right)=\frac{5\sqrt6}{2}$. The perimeter is $2r+s=4\sqrt6+\frac{5\sqrt6}{2}=\frac{13\sqrt6}{2}$.

For a sector of radius $r$ and angle $\theta$ radians, $$\frac12r^2\theta=1.$$ Its perimeter is two radii plus the arc length, so $$2r+r\theta=8.$$

From the area equation, $$\theta=\frac{2}{r^2}.$$ Substitute this into the perimeter equation: $$2r+r\left(\frac{2}{r^2}\right)=8.$$ Therefore $$2r+\frac2r=8,$$ and, since $r>0$, $$r^2-4r+1=0.$$

Complete the square: $$(r-2)^2=3,$$ so $$r=2\pm\sqrt3.$$ The condition $r>2$ selects $$r=2+\sqrt3.$$

Using $\theta=\frac{2}{r^2}$, $$\theta=\frac{2}{(2+\sqrt3)^2}. $$ Since $\frac1{2+\sqrt3}=2-\sqrt3$, $$\theta=2(2-\sqrt3)^2=2(7-4\sqrt3)=14-8\sqrt3.$$

$\text{Area of sector }BDE=\dfrac12(6^2)(1.3)=23.4\text{ m}^2$.

$12.2^2=10.8^2+6^2-2(10.8)(6)\cos\angle ABE$.

$\cos\angle ABE=\dfrac{10.8^2+6^2-12.2^2}{2(10.8)(6)}$, so $\angle ABE\approx1.54$ radians.

$\text{Area }ABE=\dfrac12(10.8)(6)\sin(1.54147...)\approx32.386$.

$\angle DBC=\pi-1.3-\angle ABE\approx0.30012$ radians.

$\text{Area }BCD=\dfrac12(6\cos\angle DBC)(6\sin\angle DBC)\approx5.084$.

$23.4+32.386+5.084\approx60.9\text{ m}^2$ to 3 significant figures.

$s=r\theta$, so $0.72=OP\times0.6$. Hence $OP=1.2\text{ m}$.

The outer sector has radius $x+1.2$ and the inner sector has radius $1.2$, with angle $0.6$ radians.

$\dfrac12(x+1.2)^2(0.6)-\dfrac12(1.2)^2(0.6)=90$.

$0.3\left((x+1.2)^2-1.44\right)=90$, so $0.3x^2+0.72x-90=0$. Multiplying by $\dfrac{50}{3}$ gives $5x^2+12x-1500=0$.

$x=\dfrac{-12+\sqrt{12^2+4(5)(1500)}}{10}=\dfrac{-6+4\sqrt{471}}{5}\approx16.16$.

$\text{Perimeter}=PQ+SR+\text{arc }QR+\text{arc }PS=x+x+0.6(x+1.2)+0.72\approx43.46\text{ m}$.

The total perimeter is $43\text{ m}$ to the nearest metre.

Since $BD$ is parallel to the diameter $AE$, the two side angles at the centre are both $0.7$ radians. Therefore $\angle BOD=\pi-0.7-0.7=\pi-1.4=1.74159\ldots$, so $1.742$ radians to 4 significant figures.

With $r=3$ and $\theta=1.74159\ldots$, sector area $=\dfrac12r^2\theta=\dfrac12(3)^2(1.74159\ldots)=7.837\ldots$.

Triangle area $=\dfrac12(3)(3)\sin(1.74159\ldots)=4.434\ldots$. The segment area is $7.837\ldots-4.434\ldots=3.403\ldots$, so $3.4\text{ m}^2$.

Arc $BCD=r\theta=3(1.74159\ldots)=5.225\ldots$.

$BD=2r\sin\dfrac{\theta}{2}=6\sin(0.87079\ldots)=4.590\ldots$. Perimeter $=5.225\ldots+4.590\ldots=9.8\text{ m}$.

In triangle $AOD$, $OD=3$, $AD=8$ and $\angle AOD=2.5$ radians.

$\dfrac{\sin\angle OAD}{3}=\dfrac{\sin2.5}{8}$, so $\angle OAD=\sin^{-1}\left(\dfrac{3\sin2.5}{8}\right)=0.2263...$.

$\angle ODA=\pi-2.5-0.2263...=0.4152...$, so $\angle ODA=0.415$ radians to 3 significant figures.

The sector has radius $3$ and angle $2\pi-2.5$, so its area is $\dfrac12(3)^2(2\pi-2.5)=17.02...$.

$\text{Area of triangle }AOD=\dfrac12(3)(8)\sin0.4152...=4.84...$.

$17.02...+4.84...=21.86...$, so the area is $21.9\text{ cm}^2$ to one decimal place.

By the cosine rule, $AO^2=3^2+8^2-2(3)(8)\cos0.4152...$, so $AO\approx5.39$ cm.

Since $A$, $B$, $O$ are collinear and $OB=3$, $AB=AO-OB\approx5.39-3=2.39$ cm.

$AB+\text{major arc }BD+AD=2.39...+3(2\pi-2.5)+8=21.7\text{ cm}$ to one decimal place.

Let $OP=r$. The area of sector $POS$ is $30$, so $\dfrac12r^2(1.65)=30$.

$r^2=\dfrac{60}{1.65}$, so $r=6.030...$ m.

$OQ=OP+PQ=6.030...+2.8=8.830...$, so $OQ=8.83$ m to 3 significant figures.

The major sector $QORTQ$ has angle $2\pi-1.65$ radians.

$\text{Area}=30+\dfrac12(8.830...)^2(2\pi-1.65)=210.6...$.

The total surface area is $211\text{ m}^2$ to the nearest integer.

The inner arc $PS$ has length $6.030...(1.65)=9.95...$, and the outer major arc has length $8.830...(2\pi-1.65)=40.91...$.

The straight sides $PQ$ and $SR$ each have length $2.8$ m.

$9.95...+40.91...+2.8+2.8=56.46...$, so the perimeter is $56\text{ m}$ to 2 significant figures.

$AO=15$, $OC=OB+BC=17$, and $\angle ACO=0.6$.

$\dfrac{\sin\angle CAO}{17}=\dfrac{\sin0.6}{15}$, so $\angle CAO\approx0.6944$ radians.

$\angle COA=\pi-0.6-0.6944...=1.847...$, so $\angle COA=1.847$ radians to 3 decimal places.

The shaded sector is the major sector $AOB$, so its angle is $2\pi-1.847...$.

$\text{Sector area}=\dfrac12(15)^2(2\pi-1.847...)=499.15...$.

$\text{Area }AOC=\dfrac12(15)(17)\sin(1.847...)=122.73...$.

$499.15...+122.73...=621.88...$, so the total area is $622\text{ m}^2$ to 3 significant figures.

By the cosine rule, $AC^2=15^2+17^2-2(15)(17)\cos(1.847...)$, so $AC\approx27.57$ m.

$\text{arc }AB=15(2\pi-1.847...)=66.54...$ m.

$66.54...+27.57...=94.11...$, so the perimeter is $94.1\text{ m}$.

$\text{arc }AB=r\theta=5(1.2)=6\text{ km}$.

$OA+OB+\text{arc }AB=5+5+6=16\text{ km}$.

$\text{Area of }OAB=\dfrac12(5^2)(1.2)=15\text{ km}^2$.

Since $\text{area }R_1=3\times\text{area }R_2$, the total area is $4\times\text{area }R_2$.

$\text{area }R_2=\dfrac{15}{4}=3.75\text{ km}^2$.

$\dfrac12(5)(OP)\sin1.2=3.75$, so $OP\approx1.609$ km.

$AP^2=5^2+OP^2-2(5)(OP)\cos1.2$, so $AP\approx4.665$ km.

$AP=4.7\text{ km}$ to the nearest $0.1$ km.

For sector $DOC$, arc length $CD=9$ and angle $0.8$ radians, so $OD=\dfrac9{0.8}=11.25\text{ m}$.

$DA:AO=3:5$, so $OD=DA+AO$ represents $8$ parts. Hence $AO=\dfrac58(11.25)=7.03125\text{ m}\approx7.03\text{ m}$.

$DA=CB=OD-AO=11.25-7.03125=4.21875\text{ m}$.

The major sector $AOBEA$ has angle $2\pi-0.8$, so arc $AEB=7.03125(2\pi-0.8)$.

$\text{Perimeter}=9+2(4.21875)+7.03125(2\pi-0.8)\approx56.0\text{ m}$.

$\text{Area}=\dfrac12(11.25)^2(0.8)+\dfrac12(7.03125)^2(2\pi-0.8)\approx186\text{ m}^2$ to the nearest whole number.

Since $O,C,D$ are in a straight line, $\angle ADO=0.43$. By the sine rule, $\dfrac{14}{\sin\angle AOD}=\dfrac{6}{\sin0.43}$, so $\sin\angle AOD=\dfrac{14\sin0.43}{6}$.

The calculator gives the acute value $1.337\ldots$, but $\angle AOD$ is obtuse. Hence $\angle AOD=\pi-1.337\ldots=1.805\ldots$, so $\angle AOD=1.805$ radians.

Arc $ABC$ is the major arc, so its angle is $2\pi-1.805\ldots=4.478\ldots$ radians.

$s=r\theta=6(4.478\ldots)=26.868\ldots$, so the length of arc $ABC$ is $26.9$ m.

The sector area is $\dfrac12r^2\theta=\dfrac12(6)^2(4.478\ldots)=80.607\ldots$.

$\angle OAD=\pi-0.43-1.805\ldots=0.9066\ldots$. The triangle area is $\dfrac12(6)(14)\sin(0.9066\ldots)=33.093\ldots$. Total area $=80.607\ldots+33.093\ldots=113.7\text{ m}^2$.

Since $OC=12$ and $DC=AB=16$, $OD=4$. In right triangle $AOD$, $\cos\angle AOD=\dfrac{OD}{OA}=\dfrac4{12}=\dfrac13$. Hence $\angle AOD=\cos^{-1}\left(\dfrac13\right)=1.231$ radians to 3 d.p.

The shaded lawn uses the major sector from $A$ to $C$, so its angle is $\pi+1.231$ radians.

$AD=\sqrt{12^2-4^2}=8\sqrt2$. Area of major sector $=\frac12(12)^2(\pi+1.231)$. Area of triangle $AOD=\frac12(4)(8\sqrt2)$.

Area $=\frac12(12)^2(\pi+1.231)-\frac12(4)(8\sqrt2)=292.2\text{ m}^2$ to 1 d.p.

Perimeter $=\text{major arc}+CD+AD=12(\pi+1.231)+16+8\sqrt2=79.8\text{ m}$ to 1 d.p.

Since $OB=4$ and the height of $B$ is $1.5$, $\cos\angle AOB=\dfrac{1.5}{4}=0.375$.

$\angle AOB=\cos^{-1}(0.375)=1.186\ldots$ rad, which is $1.186$ rad to 4 significant figures.

The horizontal coordinate of $B$ is $4\sin(1.186\ldots)=3.708\ldots$, so $BC=6-3.708\ldots=2.2919\ldots$ m.

Arc $AB=r\theta=4(1.186\ldots)=4.7456\ldots$ m.

Perimeter $=OA+\text{arc }AB+BC+CD+OD=4+4.7456\ldots+2.2919\ldots+1.5+6=18.537\ldots$ m.

Area of sector $AOB=\dfrac12(4^2)(1.186\ldots)=9.491\ldots$.

Area of $OBCD$ is the trapezium area $\dfrac12(6+2.2919\ldots)(1.5)=6.2189\ldots$. Total area $=15.710\ldots=15.7\text{ m}^2$.

Area of sector $BOC$ is $\dfrac12r^2\theta$. So $40=\dfrac12r^2(2.4)=1.2r^2$, giving $r=\sqrt{\dfrac{40}{1.2}}=5.77$ m.

The triangles $OAB$ and $ODC$ are congruent, so $\angle AOB=\angle COD$. Since $AOD$ is a straight line, $2\angle AOB+2.4=\pi$.

$\angle AOB=\dfrac{\pi-2.4}{2}=0.3708\ldots$, so $0.37$ rad.

The two congruent triangle areas together are $2\left(\dfrac12(6.25)(5.7735\ldots)\sin0.3708\ldots\right)=13.075\ldots$.

Total area $=40+13.075\ldots=53.1\text{ m}^2$ to one decimal place.

Using the cosine rule, $AB^2=6.25^2+r^2-2(6.25)(r)\cos0.3708\ldots$, giving $AB=2.265\ldots$ m.

Perimeter $=AD+AB+CD+\text{arc }BC=12.5+2(2.265\ldots)+5.7735\ldots(2.4)=30.9$ m.

For arc $AB$, $s=r\theta$, so $15=9\theta$ and $\theta=\dfrac53$ radians.

Each identical section has total angle $\dfrac{2\pi}{3}$, so the angle for arc $CD$ is $\dfrac{2\pi}{3}-\dfrac53=\dfrac{2\pi-5}{3}$.

$CD=84\left(\dfrac{2\pi-5}{3}\right)=28(2\pi-5)=35.9\text{ cm}$ to 3 significant figures.

There are 3 inner arcs, 3 outer arcs, and 6 straight radial edges of length $84-9=75$. Thus $P=3(15)+3(35.929\ldots)+6(75)=602.8\ldots\text{ cm}$.

One section has area $\dfrac12(9^2)\left(\dfrac53\right)+\dfrac12(84^2)\left(\dfrac{2\pi-5}{3}\right)=1576.525\ldots\text{ cm}^2$.

The total area is $3(1576.525\ldots)=4729.57\ldots\text{ cm}^2=4.73\times10^3\text{ cm}^2$ to 3 significant figures.

In triangle $OAB$, $OA=0.6$, $AB=1.4$, and $\angle OAB=2$ radians.

By the cosine rule, $OB^2=0.6^2+1.4^2-2(0.6)(1.4)\cos2$, so $OB=1.73756\ldots$.

Using the cosine rule again, $\cos\angle AOB=\dfrac{0.6^2+OB^2-1.4^2}{2(0.6)(OB)}$, giving $\angle AOB=0.822199\ldots$ radians.

The kite is symmetric, so $\theta=2\angle AOB=1.644398\ldots$, hence $\theta=1.64$ radians to 3 significant figures.

The sector $OCXA$ uses the major angle $2\pi-\theta$. Its arc length is $0.6(2\pi-1.644398\ldots)=2.78327\ldots$.

Perimeter $=AB+CB+\text{major arc}=1.4+1.4+2.78327\ldots=5.58327\ldots$, so the perimeter is $5.6$ m.

The kite consists of two congruent triangles like $OAB$, so area of kite $=2\left(\dfrac12(0.6)(1.4)\sin2\right)=0.76381\ldots$.

Major sector area $=\dfrac12(0.6)^2(2\pi-1.644398\ldots)=0.83498\ldots$. Total area $=0.76381\ldots+0.83498\ldots=1.59879\ldots$, so the area is $1.6\text{ m}^2$.

On the y-axis, $x=0$. From $4y+3x=48$, $4y=48$, so the y-coordinate of $C$ is $12$.

$4y+3x=48$ gives $y=-\dfrac34x+12$, so the gradient of $l_1$ is $-\dfrac34$.

Since $l_2$ is perpendicular to $l_1$, its gradient is $\dfrac43$. Through $D(8,6)$, $y-6=\dfrac43(x-8)$.

$y=\dfrac43x-\dfrac{32}{3}+6=\dfrac43x-\dfrac{14}{3}$. Therefore when $x=0$, $E_y=-\dfrac{14}{3}$.

$C=(0,12)$ and $E=\left(0,-\dfrac{14}{3}\right)$, so $CE=12+\dfrac{14}{3}=\dfrac{50}{3}$.

For sector $BCE$, $r=\dfrac{50}{3}$ and $\theta=1.8=\dfrac95$. Hence arc $BE=r\theta=\dfrac{50}{3}\cdot\dfrac95=30$.

Sector area $=\dfrac12r^2\theta=\dfrac12\left(\dfrac{50}{3}\right)^2\left(\dfrac95\right)=250$.

Triangle $CDE$ has vertical base $CE=\dfrac{50}{3}$ and horizontal height $8$, so its area is $\dfrac12\cdot\dfrac{50}{3}\cdot8=\dfrac{200}{3}$. Total area $=250+\dfrac{200}{3}=\dfrac{950}{3}$.

In triangle $OAB$, $AB=1.9$, $OB=3.4$ and $\angle AOB=\dfrac\pi6$.

$\dfrac{\sin\angle OAB}{3.4}=\dfrac{\sin(\pi/6)}{1.9}$, so $\sin\angle OAB=\dfrac{17}{19}$.

Since $\angle OAB>\dfrac\pi2$, $\angle OAB=\pi-\sin^{-1}\left(\dfrac{17}{19}\right)\approx2.034$ radians.

Using the sine rule again, $OA\approx2.096$ cm.

$\text{Area }OAB=\dfrac12(OA)(OB)\sin(\pi/6)\approx1.78\text{ cm}^2$.

$\text{Area sector }OBC:\text{Area triangle }OAB=3:2$, so sector area $=\dfrac32(1.78156...)$.

$\dfrac12(3.4)^2\theta=\dfrac32(1.78156...)$, so $\theta\approx0.462$ radians.

The perimeter is $OA+AB+OC+\text{arc }BC=2.096+1.9+3.4+3.4(0.46234...)\approx8.968$ cm.

The perimeter is $9$ cm to the nearest integer.

The sector AOBX is the major sector, so its angle is $2\pi-\dfrac{2\pi}{3}=\dfrac{4\pi}{3}$.

$\text{Area}=\dfrac12r^2\theta=\dfrac12(3^2)\left(\dfrac{4\pi}{3}\right)=6\pi\text{ m}^2$.

The arc length is $r\theta=3\left(\dfrac{4\pi}{3}\right)=4\pi$. Adding the two radii gives $4\pi+6\text{ m}$.

$\text{Area}=\dfrac12(3)(3)\sin\left(\dfrac{2\pi}{3}\right)=\dfrac92\cdot\dfrac{\sqrt3}{2}=\dfrac{9\sqrt3}{4}\text{ m}^2$.

By the cosine rule, $AB^2=3^2+3^2-2(3)(3)\cos\left(\dfrac{2\pi}{3}\right)=18+9=27$, so $AB=3\sqrt3$.

In triangle ABC, $AB^2=AC^2+8^2-2(AC)(8)\cos\left(\dfrac{\pi}{6}\right)$, so $27=AC^2-8\sqrt3\,AC+64$.

Solving gives $AC=4\sqrt3\pm\sqrt{11}$. From the diagram, $AC$ is the longer value, so $AC=4\sqrt3+\sqrt{11}$.

Area of triangle ABC is $\dfrac12(8)(AC)\sin\left(\dfrac\pi6\right)=2AC=8\sqrt3+2\sqrt{11}$. Total area is $6\pi+\dfrac{9\sqrt3}{4}+8\sqrt3+2\sqrt{11}=43.236\ldots$, so $43\text{ m}^2$ to 2 significant figures.