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Trigonometry, 2ADV T1 EQ-Bank 1

A discus throwing event is held at a field in the shape of a sector of a circle, centre \(O\), as shown in the diagram below.

Two officials are positioned at points \(P\) and \(Q\), which are 80 metres apart. The length of arc \(PQ\) is 120 metres.

The radius of the sector is \(r\) metres and the angle subtended at the centre of the arc is \(2\theta\) radians.
 

  1. Show that  \(\sin \theta=\dfrac{2 \theta}{3}\).   (2 marks)

  2. If  \(\theta=\dfrac{\pi}{3}\), find the exact area of the field.   (2 marks)

Show Answers Only

a.   \(\text{See Worked Solution.}\)

b.   \(A=\dfrac{10\ 800}{\pi}\ \text{u}^2\)

Show Worked Solution

a.   \(\text{Consider arc} \ PQ:\)

\(\dfrac{2 \theta}{2 \pi} \times 2 \pi r\) \(=120\)
\(r\) \(=\dfrac{60}{\theta}\)

\(\sin \theta=\dfrac{40}{\frac{60}{\theta}}=40 \times \dfrac{\theta}{60}=\dfrac{2 \theta}{3}\)
 

b.     \(A\) \(=\dfrac{\frac{2 \pi}{3}}{2 \pi} \times \pi r^2\)
    \(=\dfrac{\pi}{3} \times\left(\dfrac{60}{\frac{\pi}{3}}\right)^2\)
    \(=\dfrac{\pi}{3} \times 3600 \times \dfrac{9}{\pi^2}\)
    \(=\dfrac{10\ 800}{\pi}\ \text{u}^2\)

Trigonometry, 2ADV T1 EQ-Bank 4

The circle below has centre \(O\), radius \(r\) and arc length \(l\).

The ratio of the radius to the length of the arc is \(2: 5\).
 

Show that the area of the sector is  \(A=\dfrac{5 r^2}{4}\).   (2 marks)

Show Answers Only

\(\dfrac{r}{l} = \dfrac{2}{5}\ \ \Rightarrow\ \ \ l=\dfrac{5r}{2} \)

\(\text{Area}\ = \dfrac{\theta}{360} \times \pi r^2\ \ …\ (1) \)

\(\text{Arc length}\ (l): \)

\(\dfrac{\theta}{360} \times 2\pi r\) \(=l\)  
\(\dfrac{\theta}{360}\) \(=\dfrac{l}{2\pi r} = \dfrac{\frac{5r}{2}}{2\pi r} = \dfrac{5}{4 \pi}\)  

 
\(\text{Substitute}\ \ \dfrac{\theta}{360}=\dfrac{5}{4 \pi}\ \ \text{into (1):}\)

\(\text{Area}\ = \dfrac{5}{4\pi} \times \pi r^2 = \dfrac{5r^2}{4}\ \ …\ \text{as required}\)

Show Worked Solution

\(\dfrac{r}{l} = \dfrac{2}{5}\ \ \Rightarrow\ \ \ l=\dfrac{5r}{2} \)

\(\text{Area}\ = \dfrac{\theta}{360} \times \pi r^2\ \ …\ (1) \)

\(\text{Arc length}\ (l): \)

\(\dfrac{\theta}{360} \times 2\pi r\) \(=l\)  
\(\dfrac{\theta}{360}\) \(=\dfrac{l}{2\pi r} = \dfrac{\frac{5r}{2}}{2\pi r} = \dfrac{5}{4 \pi}\)  

 
\(\text{Substitute}\ \ \dfrac{\theta}{360}=\dfrac{5}{4 \pi}\ \ \text{into (1):}\)

\(\text{Area}\ = \dfrac{5}{4\pi} \times \pi r^2 = \dfrac{5r^2}{4}\ \ …\ \text{as required}\)

Trigonometry, 2ADV T1 2023 HSC 16

The diagram shows a shape `APQBCD`. The shape consists of a rectangle `ABCD` with an arc `PQ` on side `AB` and with side lengths `BC` = 3.6 m and `CD` = 8.0 m.

The arc `PQ` is an arc of a circle with centre `O` and radius 2.1 m and `∠POQ=110°`.

 

What is the perimeter of the shape `APQBCD`? Give your answer correct to one decimal place.  (4 marks)

Show Answers Only

`23.8\ text{m}`

Show Worked Solution
`text{Arc}\ PQ` `=110/360 xx pi xx 2 xx 2.1`  
  `=4.03171… \ text{m}`  

 
`text{Consider}\ ΔOPQ:`
 
 

`sin 55^@` `=x/2.1`  
`x` `=2.1 xx sin 55^@`  
  `=1.7202…`  

 
`PQ=2x=3.440\ text{m}`

`:.\ text{Perimeter}` `=8+(2xx3.6)+4.031+(8-3.440)`  
  `=23.79…\ text{m}`  
  `=23.8\ text{m  (to 1 d.p.)}`  

Trigonometry, 2ADV T1 2021 HSC 12

A right-angled triangle  `XYZ`  is cut out from a semicircle with centre `O`. The length of the diameter  `XZ`  is 16 cm and  `angle YXZ`  = 30°, as shown on the diagram.
 


 

  1. Find the length of  `XY`  in centimetres, correct to two decimal places.  (2 marks)

  2. Hence, find the area of the shaded region in square centimetres, correct to one decimal place.  (3 marks)

Show Answers Only
  1. `13.86 \ text{cm}`
  2. `45.1 \ text{cm}^2`
Show Worked Solution

 

a.    `cos 30^@` `=(XY)/16`
  `XY` `= 16 \ cos 30^@`
    `= 13.8564`
    `= 13.86 \ text{cm (2 d.p.)}`

 

b.    `text{Area of semi-circle}` `= 1/2 times pi r^2`
    `= 1/2 pi times 8^2`
    `= 100.531 \ text{cm}^2`

 

`text{Area of} \ Δ XYZ` `= 1/2 a b sin C`  
  `= 1/2 xx 16 xx 13.856 xx sin 30^@`  
  `= 55.42 \ text{cm}^2`  

 

`:. \ text{Shaded Area}` `= 100.531 – 55.42`  
  `= 45.111`  
  `= 45.1 \ text{cm}^2 \ text{(1 d.p.)}`  

Trigonometry, 2ADV T1 2020 HSC 22

The diagram shows a regular decagon (ten-sided shape with all sides equal and all interior angles equal). The decagon has centre `O`.
 

The perimeter of the shape is 80 cm.

By considering triangle `OAB`, calculate the area of the ten-sided shape. Give your answer in square centimetres correct to one decimal place.  (4 marks)

Show Answers Only

`492.4\ text(cm²)`

Show Worked Solution

`angle AOB = 360/10 = 36^@`

`AB = 80/10 = 8\ text(cm)`

`DeltaAOB\ text(is made up of 2 identical right-angled triangles)`
 

`tan 18^@` `= 4/x`
`x` `= 4/(tan 18^@)`

 

`:.\ text(Area of decagon)` `= 20 xx 1/2 xx 4/(tan 18^@) xx 4`
  `= 492.429…`
  `= 492.4\ text(cm²  (to 1 d.p.))`

Trigonometry, 2ADV T1 2019 HSC 13b

The diagram shows a circle with centre `O` and radius 20 cm.

The points `A` and `B` lie on the circle such that  `∠AOB = 70^@`.
 


 

Find the perimeter of the shaded segment, giving your answer correct to one decimal place.  (3 marks)

Show Answers Only

`47.4\ text{cm}`

Show Worked Solution
`text(Arc)\ AB` `= 70/360 xx 2pi xx 20`
  `= 24.43…`

 
`text(Using cosine rule,)`

`AB^2` `= 20^2 xx 20^2 – 2 ⋅ 20 ⋅ 20 xx cos 70`
  `= 526.383…`
`AB` `= 22.94…`

 

`:.\ text(Perimeter)` `= 24.43 + 22.94…`
  `= 47.37`
  `= 47.4\ text{cm (1 d.p.)}`

Trigonometry, 2ADV T1 SM-Bank 1 MC

A windscreen wiper blade can clean a large area of windscreen glass, as shown by the shaded area in the diagram below.
 


 

The windscreen wiper blade is 30 cm long and it is attached to a 9 cm long arm.

The arm and blade move back and forth in a circular arc with an angle of 110° at the centre.

The area cleaned by this blade, in square centimetres, is closest to

  1.    786
  2.  1382
  3.  2573
  4.  4524
Show Answers Only

`B`

Show Worked Solution
`text(Area cleaned)` `=\ text(large sector − small sector)`
  `= 110/360 xx pi xx 39^2 – 110/360 xx pi xx 9^2`
  `= 1382.3…\ \ text(cm²)`

`=>B`

Trigonometry, 2ADV T1 2017 HSC 11e

In the diagram, `OAB` is a sector of the circle with centre `O` and radius 6 cm, where  `/_ AOB = 30^@`.


 

  1.  Find the exact value of the area of the triangle `OAB`.  (1 mark)

  2. Find the exact value of the area of the shaded segment.  (1 mark)

Show Answers Only
  1.  `9\ text(cm²)`
  2. `3 pi – 9\ text(cm²)`
Show Worked Solution
i.   `text(Area)\ Delta OAB` `= 1/2 ab sin C`
    `= 1/2 xx 6^2 xx sin 30^@`
    `= 9\ text(cm²)`

 

ii.  `text(Area segment)` `= text(Area sector) – text(Area)\ Delta OAB`
    `= 30/360 xx pi xx 6^2 – 9`
    `= 3 pi – 9\ \ text(cm²)`

Trigonometry, 2ADV T1 2016 HSC 7 MC

The circle centred at `O` has radius 5. Arc  `AB`  has length 7 as shown in the diagram.
 

 hsc-2016-7mc

 
What is the area of the shaded sector `OAB?`

  1. `35/2`
  2. `35/2 pi`
  3. `125/14`
  4. `125/14 pi`
Show Answers Only

`A`

Show Worked Solution
`text(Arc)` `= theta/(2 pi) xx 2 pi r = r theta`
`7` `= 5 theta`
`:. theta` `= 7/5`

 

`text(Area of shaded sector)`

`= 1/2 theta r^2`

`= 1/2 xx 7/5 xx 5^2`

`= 35/2\ text(u²)`

`=>  A`

Trigonometry, 2ADV T1 2007 HSC 4c

 
An advertising logo is formed from two circles, which intersect as shown in the diagram.

The circles intersect at `A` and `B` and have centres at `O` and `C`.

The radius of the circle centred at `O` is 1 metre and the radius of the circle centred at `C` is `sqrt 3` metres. The length of `OC` is 2 metres.

  1. Use Pythagoras’ theorem to show that  `/_OAC = pi/2`.  (1 mark)

  2. Find  `/_ ACO`  and  `/_ AOC`.  (2 marks)

  3. Find the area of the quadrilateral  `AOBC`.  (1 mark)

  4. Find the area of the major sector  `ACB`.  (1 mark)

  5. Find the total area of the logo (the sum of all the shaded areas).  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `/_ ACO = pi/6\ ,\ /_ AOC = pi/3`
  3. `sqrt 3\ \ text(m²)`
  4. `(5 pi)/2\ text(m²)`
  5. `((19 pi + 6 sqrt 3)/6)\ text(m²)`
Show Worked Solution

i.

`text(In)\ Delta AOC`

`AO^2 + AC^2` `= 1^2 + sqrt 3^2`
  `=1 + 3`
  `= 4`
  `= OC^2`

 
`:. Delta AOC\ \ text(is right-angled and)\ \ /_OAC = pi/2`

 

ii.  `sin\  /_ACO` `= 1/2`
`:. /_ACO` `= pi/6`
`sin\  /_AOC` `= sqrt 3/2`
`:. /_AOC` `= pi/3`

 

iii.  `text(Area)\ AOBC`

`= 2 xx text(Area)\ Delta AOC`

`= 2 xx 1/2 xx b xx h`

`= 2 xx 1/2 xx 1 xx sqrt 3`

`= sqrt 3\ \ text(m²)`

 

iv.  `/_ACB = pi/6 + pi/6 = pi/3`

`:. /_ACB\ text{(reflex)}` `= 2 pi – pi/3`
  `= (5 pi)/3`

 
`text(Area of major sector)\ ACB`

`= theta/(2 pi) xx pi r^2`

`= {(5 pi)/3}/(2 pi) xx pi(sqrt 3)^2`

`= (5 pi)/6 xx 3`

`= (5 pi)/2\ text(m²)`

 

v.  `/_AOB = pi/3 + pi/3 = (2 pi)/3`

`:. /_AOB\ text{(reflex)}` `= 2 pi – (2 pi)/3`
  `= (4 pi)/3`

`text(Area of major sector)\ AOB`

`= {(4 pi)/3}/(2 pi) xx pi xx 1^2`

`= (2 pi)/3\ text(m²)`

 

`:.\ text(Total area of the logo)`

`= (5 pi)/2 + (2 pi)/3 + text(Area)\ AOBC`

`= (15 pi + 4 pi)/6 + sqrt 3`

`= ((19 pi + 6 sqrt 3)/6)\ text(m²)`

Trigonometry, 2ADV T1 2004 HSC 4a

2004 4a
 

`AOB`  is a sector of a circle, centre  `O`  and radius 6 cm.

The length of the arc  `AB`  is  `5pi` cm.

Calculate the exact area of the sector  `AOB`.   (2 marks)

Show Answers Only

`15 pi\ text(cm²)`

Show Worked Solution
`text(Arc length)\ ` `= theta/(2 pi) xx 2 pi r = r theta`
`5 pi` `= 6 theta`
`:.\ theta` `= (5pi)/6\ text(radians)`

 

`text(Area of sector)\ AOB`

`= theta/(2pi) xx pi r^2`

`= 1/2 r^2 theta`

`= 1/2 xx 6^2 xx (5pi)/6`

`= 15pi\ text(cm²)`

Trigonometry, 2ADV T1 2006 HSC 4a

In the diagram, `ABCD` represents a garden. The sector  `BCD`  has centre `B` and  `/_DBC = (5 pi)/6`

The points `A, B` and `C` lie on a straight line and  `AB = AD = 3` metres.

Copy or trace the diagram into your writing booklet.

  1. Show that  `/_DAB = (2 pi)/3.`  (1 mark)

  2. Find the length of  `BD`.  (2 marks)

  3. Find the area of the garden  `ABCD`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `3 sqrt 3\ \ text(m)`
  3. `(9 sqrt 3 + 45 pi) / 4\ \ text(m²)`
Show Worked Solution
i.   

`text(Show)\ /_DAB = (2 pi)/3`

`/_DBA` `= pi – (5 pi)/6\ \ \ text{(π radians in straight angle}\ ABC text{)}`
  `= pi/6\ text(radians)`

 
`:. /_BDA = pi/6\ text(radians)\ \ \ text{(base angles of isosceles}\ Delta ADB text{)}`

`:. /_DAB` `= pi – (pi/6 + pi/6)\ \ \ text{(angle sum of}\ Delta ADB text{)}`
  `= (2 pi)/3\  text(radians … as required)`

 

ii.  `text(Using the cosine rule:)`

`BD^2` `= AD^2 + AB^2 – 2 xx AD xx AB xx cos {:(2 pi)/3`
  `= 9 + 9 – (2 xx 3 xx 3 xx -0.5)`
  `= 27`
`:. BD` `= sqrt 27`
  `= 3 sqrt 3\ \ text(m)`

 

iii.  `text(Area of)\ Delta ADB` `= 1/2 ab sin C`
  `= 1/2 xx 3 xx 3 xx sin{:(2 pi)/3`
  `= 9/2 xx sqrt3/2`
  `= (9 sqrt 3)/4\ \ text(m²)`

 
`text(Area of sector)\ BCD`

`= {(5 pi)/6}/(2 pi) xx pi r^2`

`= (5 pi)/12 xx (3 sqrt 3)^2`

`= (45 pi)/4\ \ text(m²)`

 

`:.\ text(Area of garden)\ ABCD`

`= (9 sqrt 3)/4 + (45 pi)/4`

`= (9 sqrt 3 + 45 pi)/4\ \ text(m²)`

Trigonometry, 2ADV T1 2005 HSC 4a

 Trig Calculus, 2UA 2005 HSC 4a
 

A pendulum is 90 cm long and swings through an angle of 0.6 radians. The extreme positions of the pendulum are indicated by the points `A` and `B` in the diagram.

  1. Find the length of the arc  `AB`.  (1 mark)

  2. Find the straight-line distance between the extreme positions of the pendulum.  (2 marks)

  3. Find the area of the sector swept out by the pendulum.  (1 mark)

Show Answers Only
  1. `text(54 cm)`
  2. `text{53.2 cm  (to 1 d.p.)}`
  3. `text(2430 cm²)`
Show Worked Solution
i.    `text(Arc)\ AB` `= theta/(2 pi) xx 2 pi r`
    `= rtheta`
    `= 90 xx 0.6`
    `= 54\ text(cm)`

 

ii. 

 Trig Calculus, 2UA 2005 HSC 4a Answer

`text(Using the cosine rule)`

`text(Distance)\ AB\ text(in straight line)`

`AB^2` `= 90^2 xx 90^2 − 2 xx 90 xx 90 xx cos\ 0.6`
  `= 2829.563…`
`:.AB` `= 53.193…`
  `= 53.2\ text{cm  (to 1 d.p.)}`

 

iii. `text(Area of Sector)`

`= 0.6/(2pi) xx pir^2`

`= 0.3 xx 90^2`

`= 2430\ text(cm²)`

Trigonometry, 2ADV T1 2008 HSC 7b

2008 7b

The diagram shows a sector with radius  `r`  and angle  `theta`  where  `0 < theta <= 2pi`.

The arc length is  `(10pi)/3`. 

  1.  Show that  `r >= 5/3`.   (2 marks)

  2.  Calculate the area of the sector when  `r = 4`.    (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `(20pi)/3\ text(u²)`
Show Worked Solution
i.    `text(Show)\ r >= 5/3`
`text(Arc length)\ ` `= r theta\ \ text(where)\ \ 0 < theta <= 2pi`
`r theta` `= (10pi)/3`
`:.theta` `= (10pi)/(3r)`

 

`text(Using)\ \ \ 0 <= theta <= 2 pi`

`0 <= (10pi)/(3r)` `<= 2pi`
`(10pi)/3` `<= 2 pi r`
`5/3` `<= r`

 

`:.\ r >= 5/3\ \ \ text(… as required.)`

 

ii.   `text(Area)` `= 1/2 r^2 theta`
    `= 1/2 xx 4^2 xx (10pi)/(3 xx 4)`
    `= (20pi)/3\ text(u²)`

Trigonometry, 2ADV T1 2014 HSC 11g

The angle of a sector in a circle of radius 8 cm is  `pi/7`  radians, as shown in the diagram.  
  

2014 11g

 
Find the exact value of the perimeter of the sector.   (2 marks)

Show Answers Only

`(8pi)/7 + 16\ text(cm)`

Show Worked Solution
`text(Arc length)` `= theta/(2pi) xx 2 pi r`
  `= pi/7 xx 8`
  `= (8pi)/7\ text(cm)`

 

`text(S)text(ector perimeter)` `= text(arc) + 2 xx text(radius)`
  `= (8pi)/7 + 16\ text(cm)`

Trigonometry, 2ADV T1 2009 HSC 5c

The diagram shows a circle with centre `O` and radius 2 centimetres. The points  `A`  and  `B`  lie on the circumference of the circle and  `/_AOB = theta`.
 

2009 5c  
 

  1. There are two possible values of  `theta`  for which the area of  `Delta AOB`  is  `sqrt 3`  square centimetres. One value is  `pi/3`.

     

    Find the other value.    (2 marks)

  2. Suppose that  `theta = pi/3`.

     

    (1)  Find the area of sector  `AOB`   (1 mark)

     

    (2)  Find the exact length of the perimeter of the minor segment bounded by the chord  `AB`  and the arc  `AB`.   (2 marks)

Show Answers Only
  1. `(2pi)/3`
  2. (1)  `(2pi)/3\ \ text(cm²)`
  3. (2)  `(2 + (2pi)/3)\ text(cm)`
Show Worked Solution
i.    `text(Area)\ Delta AOB` `= 1/2 ab sin theta`
    `= 1/2 xx 2 xx 2 xx sin theta`
    `= 2 sin theta`
`2 sin theta` `= sqrt 3\ \ \ text{(given)}`
`sin theta` `= sqrt3/2`
`:. theta` `= pi/3,\ pi\ – pi/3`
  `= pi/3,\ (2pi)/3`

 

`:.\ text(The other value of)\ theta\ text(is)\ \ (2pi)/3\ \ text(radians)` 

 

ii. (1)    `text(Area of sector)\ AOB` `= pi r^2 xx theta/(2pi)`
    `= 1/2 r^2 theta`
    `= 1/2 xx 2^2 xx pi/3`
    `= (2pi)/3\ text(cm²)` 

 

ii. (2)    `text(Using the cosine rule:)`
`AB^2` `= OA^2 + OB^2\ – 2 xx OA xx OB xx cos theta`
  `= 2^2 + 2^2\ – 2 xx 2 xx 2 xx cos (pi/3)`
  `= 4 + 4\ – 4`
  `= 4`
`:.\ AB` `= 2`

 

`text(Arc)\ AB` `= 2 pi r xx theta/(2pi)`
  `= r theta`
  `= (2pi)/3\ text(cm)`

 

`:.\ text(Perimeter) = (2 + (2pi)/3)\ text(cm)`

Trigonometry, 2ADV T1 2010 HSC 6b

The diagram shows a circle with centre `O` and radius 5 cm.

The length of the arc `PQ` is 9 cm. Lines drawn perpendicular to `OP` and `OQ` at `P` and `Q` respectively meet at  `T`.
 

  1. Find  `/_POQ`  in radians.   (1 mark)

  2. Prove that  `Delta OPT`  is congruent to  `Delta OQT`.   (2 marks)

  3. Find the area of the shaded region.   (2 marks)

Show Answers Only
  1. `9/5\ text(radians)`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `6.3\ text(cm)\ \ \ text{(to 1 d.p.)}`
  4. `9.0\ text(cm²)\ \ \ text{(to 1 d.p.)}`
Show Worked Solution
i.    `text(Length of Arc)` `= r theta`
  `9` `= 5 xx /_POQ`
  `:.\ /_ POQ` `= 9/5\ text(radians)`

 

ii.

`text(Prove)\ Delta OPT ~= Delta OQT`

`OT\ text(is common)`

 MARKER’S COMMENT: Know the difference between the congruency proof of `RHS` and `SAS`. Incorrect identification will lose a mark.

`/_OPT = /_OQT = 90°\ \ \ text{(given)}`

`OP = OQ\ \ \ text{(radii)}`

`:.\ Delta OPT ~= Delta OQT\ \ \ text{(RHS)}`
 

iii.   
`/_POT ` `= 1/2 xx /_POQ\ \ \ text{(from part (ii))}`
  `=1/2 xx 9/5`
  `= 9/10\ text(radians)`
♦♦ Mean mark below 30%.
MARKER’S COMMENT: Many students struggled to work in radians. Make sure you understand this concept.
`tan /_ POT` `= (PT)/(OP)`
`tan (9/10)` `= (PT)/5`
`PT` `= 5 xx tan(9/10)`
  `=6.3007…`
  `=6.3\ text(cm)\ \ text{(to 1 d.p.)}`

 

iv.    `text(Shaded Area = Area)\ OQTP\ – text(Area Sector)\ OQP`
♦ Mean mark 35%.
`text(Area)\ OQTP` `= 2 xx text(Area)\ Delta OPT`
  `=2 xx 1/2 xx OP xx PT`
  `= 5 xx 6.3007`
  `~~ 31.503…`
  `~~31.5\ text(cm²)`

 

`text(Area Sector)\ OQP` ` = 1/2 r^2 theta`
  `= 1/2 xx 25 xx 9/5`
  `= 22.5\ text(cm²)`

 

`:.\ text(Shaded Area)` `= 31.503\ – 22.5`
  `=9.003…`
  `=9.0\ text(cm²)\ \ \ text{(to 1 d.p.)}`

Trigonometry, 2ADV T1 2012 HSC 11f

The area of the sector of a circle with a radius of 6 cm is 50 cm².

Find the length of the arc of the sector.  (2 marks)

Show Answers Only

`50/3\ text(cm)`

Show Worked Solution
TIP: Many students find it easier to think of the area of a sector by calculating `theta/(2 pi)` multiplied by the area of a circle rather than remembering a formula.

`text(Area of sector, radius 6 cm = 50 cm²)`

`theta/(2 pi) xx pi r^2` `= 50`
`1/2 r^2 theta` `=50`
`1/2 xx 6^2 xx theta`  `=50`
`theta` `=50/18=25/9\ text(radians)`

 

`:.\ text(Length of Arc)` `= theta/(2pi) xx 2pi r` 
  `= theta xx r`
  `= 25/9xx6` 
  `= 50/3 \ text(cm)`

 

Trigonometry, 2ADV T1 2013 HSC 13c

The region  `ABC`  is a sector of a circle with radius 30 cm, centred at  `C`. The angle of the sector is  `theta`. The arc  `DE`  lies on a circle also centred at  `C`, as shown in the diagram.
 

2013 13c

The arc  `DE`  divides the sector  `ABC`  into two regions of equal area.

Find the exact length of the interval  `CD`.   (2 marks)

Show Answers Only

 `15sqrt2\ text(cm.)`

Show Worked Solution
`text(Area of sector)\ ABC` `=1/2 r^2 theta`
  `=1/2 xx 30^2 xx theta`
  `= 450 theta`

`text(Let)\ CD = x`

MARKER’S COMMENT: Simply finding the area of sector `ABC` achieved half marks in this challenging question! Show your working.

`text(Area of sector)\ CDE = 1/2 x^2 theta`

`text(S)text(ince)\ DE\ text(divides sector)\ ABC\ text(in half,)`

`text(Area sector)\ CDE` `= 1/2 xx text(Area sector)\ ABC`
`1/2 x^2 theta` `= 1/2 xx 450 theta`
`x^2` `=450`
`x` `=sqrt450\ \ \ \ \ \ (x>0)`
  `= 15 sqrt2\ text(cm)`

 

`:.\ text(The exact length of interval)\ CD\ text(is)\ 15 sqrt2\ text(cm.)`