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Trigonometry, 2ADV T1 EQ-Bank 6 MC

The triangle \(A B C\) has an exact area of \((12-\sqrt{3})\) cm\(^{2}\).
 

The exact value of \(x\) is

  1. \(\dfrac{24-2\sqrt{3}}{3}\)
  2. \(\dfrac{8\sqrt{3}-2}{3} \)
  3. \(8\sqrt{3}-1\)
  4. \(12\sqrt{3}+1\)
Show Answers Only

\(B\)

Show Worked Solution
  \(12-\sqrt{3}\) \(=\dfrac{1}{2} \cdot 6 \cdot x \cdot \sin 120^{\circ}\)
  \(12-\sqrt{3}\) \(=3x \cdot \sin 120^{\circ}\)
  \(12-\sqrt{3}\) \(=\dfrac{3\sqrt{3}x}{2}\)
  \(x\) \(=\dfrac{2(12-\sqrt{3})}{3\sqrt{3}} \times \dfrac {\sqrt{3}}{\sqrt{3}}\)
    \(=\dfrac{24\sqrt{3}-6}{9} \)
    \(=\dfrac{8\sqrt{3}-2}{3} \)

 
\(\Rightarrow B\)

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 2021 HSC 18

The diagram shows a triangle `ABC` where `AC` = 25 cm, `BC` = 16 cm, `angle BAC` = 28° and angle `ABC` is obtuse.
 


 

Find the size of the obtuse angle `ABC` correct to the nearest degree.  (3 marks)

Show Answers Only

`133°`

Show Worked Solution

`text(Using the sine rule:)`

`sin theta/25` `= (sin 28°)/16`
`sin theta` `= (25 xx sin 28°)/16`
`sin theta` `= 0.73355`
`theta` `= 47°`
 
`:. angleABC` `= 180-47`
  `= 133°`

Trigonometry, 2ADV T1 EQ-Bank 2

Determine all possible dimensions for triangle  `ABC`  given  `AB = 6.2\ text(cm)`, `angleABC = 35°`  and  `AC = 4.1`.

Give all dimensions correct to one decimal place.  (3 marks)

Show Answers Only

`text(7.1 cm, 6.2 cm, 4.1 cm or)`

`text(3.0 cm, 6.2 cm, 4.1 cm.)`

Show Worked Solution

`text(Using the sine rule:)`

`(sinangleACB)/6.2` `= (sin35^@)/4.1`
`sinangleACB` `= (6.2 xx sin35^@)/4.1`
  `= 0.8673…`
`angleACB` `= 60.15…^@\ text(or)\ 119.84…^@`

  
`text(If)\ \ angleACB = 60.15^@,`

`angleBAC = 180 – (35 + 60.15) = 84.85^@`
 

`(BC)/(sin84.85)` `= 4.1/(sin35^@)`
`BC` `= 7.11…`
  `= 7.1\ text(cm)`

 
`text(If)\ \ angleACB = 119.85^@,`

`angleBAC = 180 – (35 + 119.85) = 25.15^@`
 

`(BC)/(sin25.15)` `= 4.1/(sin35^@)`
`BC` `= 3.03…`
  `= 3.0\ text(cm)`

 
`:.\ text(Possible dimensions are:)`

`text(7.1 cm, 6.2 cm, 4.1 cm or)`

`text(3.0 cm, 6.2 cm, 4.1 cm.)`

Trigonometry, 2ADV T1 2018 HSC 14a

In  `Delta KLM, KL`  has length 3, `LM` has length 6 and `/_KLM` is 60°. The point `N` is chosen on side  `KM`  so that  `LN`  bisects `/_KLM`. The length  `LN`  is `x`.
 


 

  1. Find the exact value of the area of  `Delta KLM`.  (1 mark)

  2. Hence, or otherwise, find the exact value of `x`.  (2 marks)

Show Answers Only
  1. `(9 sqrt 3)/2`
  2. `2 sqrt 3`
Show Worked Solution

i.   `text(Using sine rule:)`

`text(Area)\ \ Delta KLM` `= 1/2 xx 3 xx 6 xx sin 60^@`
  `= (9 sqrt 3)/2\ \ text(u²)`

 

ii.  `text(Area)\ \ Delta KLN + text(Area)\ \ Delta NLM = text(Area)\ \ Delta KLM`

`1/2 xx 3 xx x xx sin 30^@ + 1/2 xx x xx 6 xx sin 30^@ = (9 sqrt 3)/2`

♦ Mean mark 37%.

`3/4 x + 3/2 x` `= (9 sqrt 3)/2`
`9/4 x` `= (9 sqrt 3)/2`
`:. x` `= (9 sqrt 3)/2 xx 4/9`
  `= 2 sqrt 3`

Trigonometry, 2ADV T1 2015 HSC 13a

The diagram shows `Delta ABC` with sides  `AB = 6` cm, `BC = 4` cm  and  `AC = 8` cm.
 

  1. Show that  `cos A = 7/8`.  (1 mark)

  2. By finding the exact value of `sin A`, determine the exact value of the area of  `Delta ABC`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `3 sqrt 15\ \ text(cm²)`
Show Worked Solution

i.  `text(Show)\ cos A = 7/8`

`text(Using the cosine rule)`

`cos A` `= (b^2 + c^2-a^2)/(2bc)`
  `= (8^2 + 6^2-4^2)/(2 xx 8 xx 6)`
  `= (64 + 36-16)/96`
  `= 84/96`
  `= 7/8\ \ text(…  as required)`

 

♦ Mean mark 40%.
ii.    2UA HSC 2015 13ai
`a^2 + 7^2` `= 8^2`
`a^2 + 49` `= 64`
`a^2` `= 15`
`a` `= sqrt 15`
`:.\ sin A` `= (sqrt 15)/8`

 

`:.\ text(Area)\ Delta ABC` `= 1/2 bc\ sin A`
  `= 1/2 xx 8 xx 6 xx (sqrt 15)/8`
  `= 3 sqrt 15\ \ text(cm²)`

Trigonometry, 2ADV T1 2006 HSC 4a

In the diagram, `ABCD` represents a garden. The sector  `BCD`  has centre `B` and  `/_DBC = (5 pi)/6`

The points `A, B` and `C` lie on a straight line and  `AB = AD = 3` metres.

Copy or trace the diagram into your writing booklet.

  1. Show that  `/_DAB = (2 pi)/3.`  (1 mark)

  2. Find the length of  `BD`.  (2 marks)

  3. Find the area of the garden  `ABCD`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `3 sqrt 3\ \ text(m)`
  3. `(9 sqrt 3 + 45 pi) / 4\ \ text(m²)`
Show Worked Solution
i.   

`text(Show)\ /_DAB = (2 pi)/3`

`/_DBA` `= pi – (5 pi)/6\ \ \ text{(π radians in straight angle}\ ABC text{)}`
  `= pi/6\ text(radians)`

 
`:. /_BDA = pi/6\ text(radians)\ \ \ text{(base angles of isosceles}\ Delta ADB text{)}`

`:. /_DAB` `= pi – (pi/6 + pi/6)\ \ \ text{(angle sum of}\ Delta ADB text{)}`
  `= (2 pi)/3\  text(radians … as required)`

 

ii.  `text(Using the cosine rule:)`

`BD^2` `= AD^2 + AB^2 – 2 xx AD xx AB xx cos {:(2 pi)/3`
  `= 9 + 9 – (2 xx 3 xx 3 xx -0.5)`
  `= 27`
`:. BD` `= sqrt 27`
  `= 3 sqrt 3\ \ text(m)`

 

iii.  `text(Area of)\ Delta ADB` `= 1/2 ab sin C`
  `= 1/2 xx 3 xx 3 xx sin{:(2 pi)/3`
  `= 9/2 xx sqrt3/2`
  `= (9 sqrt 3)/4\ \ text(m²)`

 
`text(Area of sector)\ BCD`

`= {(5 pi)/6}/(2 pi) xx pi r^2`

`= (5 pi)/12 xx (3 sqrt 3)^2`

`= (45 pi)/4\ \ text(m²)`

 

`:.\ text(Area of garden)\ ABCD`

`= (9 sqrt 3)/4 + (45 pi)/4`

`= (9 sqrt 3 + 45 pi)/4\ \ text(m²)`

Trigonometry, 2ADV T1 2006 HSC 1d

2006 1d

 
Find the value of `theta` in the diagram. Give your answer to the nearest degree.  (2 marks)

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`18°`

Show Worked Solution

`text(Using the sine rule)`

`sin theta / 5` `= (sin 33°)/9`
`sin theta` `= (5 xx sin 33°)/9`
  `= 0.30257…`
`:. theta` `= 17.612…`
  `= text{18°  (nearest degree)}`

Trigonometry, 2ADV T1 2005 HSC 3b

The lengths of the sides of a triangle are 7 cm, 8 cm and 13 cm.

  1. Find the size of the angle opposite the longest side.  (2 marks)

  2. Find the area of the triangle.  (1 marks)

Show Answers Only
  1. `120^@`
  2. `14sqrt3\ text(cm)`
Show Worked Solution

i.

 Trig Ratios, 2UA 2005 HSC 3b Answer1 

`∠ABC\ \ text(is opposite the longest side)`

`text(Using the cosine rule)`

`cos\ ∠ABC` `= (7^2 + 8^2 −13^2)/(2 xx 7 xx 8)`
  `= text(−)1/2`

 
`text(S)text(ince cos)\ 60^@ = 1/2\ text(and cos is negative)`

`text(in 2nd quadrant,)`

`∠ABC` `= 180− 60`
  `= 120^@`

 

ii.  `text(Using the sine rule)`

`text(Area)\ ΔABC` `= 1/2\ ab\ sin\ C`
  `= 1/2 xx 7 xx 8\ sin 120^@`
  `= 28 xx sqrt3/2`
  `= 14sqrt3\ text(cm)^2`

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)

Show Answers Only
  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 14c

2013 14c

 
The right-angled triangle  `ABC`  has hypotenuse  `AB = 13`. The point  `D`  is on  `AC`  such that  `DC = 4`,  `/_DBC = pi/6` and  `/_ABD = x`.

Using the sine rule, or otherwise, find the exact value of  `sin x`.   (3 marks)

Show Answers Only

 `(7sqrt3)/26 text(.)`

Show Worked Solution

`text(Find)\ \ /_ADB`

`/_ADB` `= pi/6 + pi/2 \ \ \ text{(exterior angle of}\ Delta BDC text{)}`
  `= (2pi)/3\ text(radians)`

 
`text(Find)\ \ AD`

♦ Mean mark 36%.
STRATEGY TIP: The hint to use the sine rule should flag to students that they will be dealing in non-right angled trig (i.e. `Delta ABD`) and to direct their energies at initially finding `/_ADB` and `AD`.
`tan (pi/6)` `= 4/(BC)`
`1/sqrt3` `=4/(BC)`
`BC` `=4 sqrt3`

 

`text(Using Pythagoras:)`

`AC^2 + BC^2` `= AB^2`
`AC^2 + (4sqrt3)^2` `= 13^2`
`AC^2` `= 169\-48`
  `= 121`
`=>AC` `= 11`
`:.AD` `=AC\-DC`
  `= 11 -4`
  `=7`

 

`text(Using sine rule:)`

`(AB)/(sin /_BDA)` `= (AD)/(sinx)`
`13/(sin ((2pi)/3))` `=7/(sinx)`
`13 xx sinx` `= 7 xx sin ((2pi)/3)`
`sinx` `= 7/13 xx sin((2pi)/3)`
  `= 7/13 xx sqrt3/2`
  `= (7 sqrt3)/26`

 
`:.\ text(The exact value of)\ sinx = (7sqrt3)/26 text(.)`