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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`
  `=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 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°  \ text{(nearest degree)}`
  3. `N (3,2)`
Show Worked Solution

i.   `text(Find distance) \  AB:`

`text(Find A),\ \ y=0`

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

 
`text(Find B),\ \ x=0`

` 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(Using 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 = ½ \ \ \ (m_1 m_2 = -1\ \ text(for ⊥ lines))`

 

`text(Equation of) \ CN, \ m = ½\ text(through)\ (7,4)`

MARKER’S COMMENT: Many students could not find the correct equation on `CN` because they took its gradient to be the reciprocal of `AB` and not the negative reciprocal. 
`y  – 4` `= ½ (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)`