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Trigonometry, 2ADV T1 2025 HSC 28

A farmer wants to use a straight fence to divide a circular paddock of radius 10 metres into two segments. The smaller segment is \(\dfrac{1}{4}\) of the paddock and is shaded in the diagram. The fence subtends an angle of \(\theta\) radians at the centre of the circle as shown.
 

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

  2. The graph of  \(y=\sin \theta+\dfrac{\pi}{2}\) is shown.
     
  4. Use the graph and the result in part (a) to estimate the arc length of the smaller segment to the nearest metre.   (2 marks)

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a.   \(\text{See Worked Solutions}\)

b.   \(\text{Arc length} \ \approx 23 \ \text{metres.}\)

Show Worked Solution

a.    \(\text{Area of paddock} =\pi \times 10^2=100 \pi\)

\(\text{Area of segment} =\dfrac{1}{4} \times 100 \pi =25 \pi\)

\(\text{Area of sector} =\dfrac{\theta}{2 \pi} \times 100 \pi=50 \theta\)

\(\text{Area of triangle} =\dfrac{1}{2} ab \, \sin C=50 \, \sin \theta\)

\(\text{Equating sector areas:}\)

\(25 \pi\) \(=50 \theta-50\, \sin \theta\)
\(50 \theta\) \(=25 \pi+50 \, \sin \theta\)
\(\theta\) \(=\dfrac{\pi}{2}+\sin \theta\)
♦♦ Mean mark (a) 34%.

b.    \(\text{Find where} \ \ \theta=\sin \theta+\dfrac{\pi}{2}\)

\(\text {Intersection occurs where: }\)

\(y=\theta \ \ \text{intersects with} \ \ y=\sin \theta+\dfrac{\pi}{2}\)

\(\text{At intersection (from graph):} \ \ \theta \approx 2.3 \ \text{radians}\)

\(\text{Arc length} \ \approx 10 \times 2.3 \approx 23 \ \text{metres.}\)

♦♦♦ Mean mark (b) 7%.

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)

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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)

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

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

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

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

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

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

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 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)

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  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 EQ-Bank 4 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
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`B`

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

`=>B`

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`
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`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 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)

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`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 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)

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  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)

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  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 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)

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  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 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)

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 `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.)`