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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)

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 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)

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  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 2005 HSC 9b

Trig Ratios, 2UA 2005 HSC 9b
 

The triangle  `ABC`  has a right angle at `B, \ ∠BAC = theta` and  `AB = 6`. The line `BD` is drawn perpendicular to `AC`. The line `DE` is then drawn perpendicular to `BC`. This process continues indefinitely as shown in the diagram.

  1. Find the length of the interval `BD`, and hence show that the length of the interval  `EF`  is  `6 sin^3\ theta`.  (2 marks)

  2. Show that the limiting sum 
     
    `qquad BD + EF + GH + ···`
     
    is given by  `6 sec\ theta tan\ theta`.  (3 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution
i.   Trig Ratios, 2UA 2005 HSC 9b Answer

`text(Show)\ EF = 6\ sin^3\ theta`

`text(In)\ ΔADB`

`sin\ theta` `= (DB)/6`
`DB` `= 6\ sin\ theta`
`∠ABD` `= 90 − theta\ \ \ text{(angle sum of}\ ΔADB)`
`:.∠DBE` `= theta\ \ \ (∠ABE\ text{is a right angle)}`

 

`text(In)\ ΔBDE:`

`sin\ theta` `= (DE)/(DB)`
  `= (DE)/(6\ sin\ theta)`
`DE` `= 6\ sin^2\ theta`
`∠BDE` `= 90 − theta\ \ \ text{(angle sum of}\ ΔDBE)`
`∠EDF` `= theta\ \ \ (∠FDB\ text{is a right angle)}`

 

`text(In)\ ΔDEF:`

`sin\ theta` `= (EF)/(DE)`
  `= (EF)/(6\ sin^2\ theta)`
`:.EF` `= 6\ sin^3\ theta\ \ …text(as required)`

 

ii. `text(Show)\ \ BD + EF + GH\ …`

`text(has limiting sum)\ =6 sec theta tan theta`

`underbrace{6\ sin\ theta + 6\ sin^3\ theta +\ …}_{text(GP where)\ \ a = 6 sin theta, \ \ r = sin^2 theta}`
 

`text(S)text(ince)\ \ 0 < theta < 90^@`

`−1` `< sin\ theta` `< 1`
`0` `< sin^2\ theta` `< 1`

 
`:. |\ r\ | < 1`
 

`:.S_∞` `= a/(1 − r)`
  `= (6\ sin\ theta)/(1 − sin^2\ theta)`
  `= (6\ sin\ theta)/(cos^2\ theta)`
  `= 6 xx 1/(cos\ theta) xx (sin\ theta)/(cos\ theta)`
  `= 6 sec\ theta\ tan\ theta\ \ …text(as required.)`

Trigonometry, 2ADV T1 2013 HSC 2 MC

The diagram shows the line  `l`.

2013 2 mc

 What is the slope of the line  `l`? 

  1. `sqrt3`  
  2. `- sqrt3`  
  3. `1/sqrt3`  
  4. `- 1/sqrt3`  
Show Answers Only

`B`

Show Worked Solution

`text(Gradient is negative)`

`text{(slopes from top left to bottom right)}`

`tan60^@ = sqrt3`

`:.\ text(Gradient is)\ – sqrt3`

`=>  B`