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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= 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(m)^2`
 

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(m)^2`
 

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(m)^2`
 

`:.\ 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(m)^2`

Trigonometry, 2ADV T1 2012 HSC 13a

The diagram shows a triangle `ABC`. The line  `2x + y = 8`  meets the `x` and `y` axes at the points `A` and `B` respectively. The point `C` has coordinates `(7, 4)`. 
 

2012 13a

  1. Calculate the distance ` AB `.   (2 marks)

  2. It is known that  `AC = 5`  and  `BC = sqrt 65 \ \ \ `(Do NOT prove this)  

     

    Calculate the size of `angle ABC` to the nearest degree.   (2 marks)

  3. The point `N` lies on `AB` such that `CN` is perpendicular to `AB`. 

     

    Find the coordinates of `N`.   (3 marks)

Show Answers Only
  1. `4 sqrt 5  \ text(units)`
  2. ` 34°`
  3. `N (3,2)`
Show Worked Solution

i.   `text(Find distance)\ AB:`

`text(Find A:)`

`2x + 0` `= 8`
`x` `= 4\ \ =>\ \ A (4,0)`

 
`text(Find B:)`

` 0 + y = 8 \ \ =>\ \ B(0,8)`
  

`text(Using Pythagoras:)`

`AB^2` `= OB^2 + OA^2= 8^2 + 4^2= 80`
`:. \ AB` `= sqrt 80= 4 sqrt 5  \ text(units)`

 

ii.   `text(Find)\ angle ABC\ \text{sing cosine rule:}`

`cos angle ABC` `=  (AB^2 + BC^2-AC^2)/(2 xx AB xx BC)`
  `= ((4 sqrt 5)^2 + (sqrt 65)^2-5^2)/(2 xx 4 sqrt 5 xx sqrt 65)` 
  `= (80 + 65-25) / (8 xx sqrt 325)`
  `= 120/(40 sqrt 13)`
  `= 3/ sqrt 13`
  `= 0.83205…`

 
`:. angle ABC= 33.690…= 34°\ \  text{(nearest degree)}`
 

iii.  `text(Find)\ N:`

`AB   text(is)  \ 2x +y = 8`

`=>  \ text(Gradient)\ AB = -2`

`:.\ text(Gradient of)\ CN = 1/2 \ \ (m_1 m_2 = -1\ \ text(for ⊥ lines))`
 

`text(Equation of)\ CN, \ m = 1/2\ text(through)\ (7,4):`

`y-4` `= 1/2(x-7)`
`2y-8` `= x-7`
`x-2y + 1` `= 0`

 

` N \ text(is intersection of)\ AB\ text(and)\ CN:`

`2x + y-8` `= 0\ \ …\ (1)`
`x-2y + 1` `= 0\ \ …\ (2)`

 
`text(Multiply)  \ (1) xx 2`

`4x +2y-16 = 0\ \ …\ (3)`

 
`text{Add (2) + (3):}`

`5x-15=0\ \ =>\ \ x=3`

 
`text(Substitute)\ x = 3\ text{into (1):}`

`2(3) + y-8 = 0\ \ =>\ \ y = 2`

`:. N (3,2)`