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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)

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`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 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)

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`47.4\ text{cm}`

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`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` `= \sqrt{526.383…}=22.94…`

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

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 cm)^2`
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(cm)^2`

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 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)

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`(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`.
  3. i.  Find the area of sector `AOB`   (1 mark)

  4. ii. 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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a.    `(2pi)/3`

b.i   `(2pi)/3\ \ text(cm)^2`

b.ii  `(2 + (2pi)/3)\ text(cm)`

Show Worked Solution

a.    `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.`  
 

b.i   `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(cm)^2`
 

b.ii  `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`
`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 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)

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`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)`