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Trigonometry, 2ADV EQ-Bank 2 MC

A tower \(B T\) has height \(h\) metres.

From point \(A\), the angle of elevation to the top of the tower is 26° as shown.
 

Which of the following is the correct expression for the length of \(A B\) ?

  1. \(h\, \tan 26^{\circ}\)
  2. \(h\, \cot 26^{\circ}\)
  3. \(h\, \sin 26^{\circ}\)
  4. \(h\, \operatorname{cosec}\, 26^{\circ}\)
Show Answers Only

\(B\)

Show Worked Solution
\(\tan 26^{\circ}\) \(=\dfrac{h}{AB}\)  
\(AB\) \(=\dfrac{h}{\tan 26^{\circ}}\)  
\(AB\) \(=h\,\cot 26^{\circ}\)  

 
\(\Rightarrow B\)

Trigonometry, 2ADV T1 EQ-Bank 3

A student is vaping behind a wall in a school assembly room.

A teacher is walking up a corridor as shown in the diagram. The wall at the end is covered by a large mirror.
 

 
The line of sight off a mirror follows the law of reflection as follows:
 

Determine the distance, \(d\) metres, at which the teacher will first be able to see the student vaping.   (4 marks)

Show Answers Only

\(d=4.0 \ \text{metres}\)

Show Worked Solution

\(\text{Label alternate angles at } d_{\text{min}}:\)

\(\tan \theta=\dfrac{x}{2} \ \  \text{(bottom left \(\Delta\))}\)

\(\tan \theta=\dfrac{1-x}{1}\ \  \text{\((\Delta\) with side  \(1-x\))}\)

\(\dfrac{x}{2}=1-x\)

\(x=2-2 x\)

\(x=\dfrac{2}{3}\)

\(\text{Consider top left triangle:}\)

\(\tan \theta\) \(=\dfrac{\frac{4}{3}}{d}\)
\(\dfrac{1}{3}\) \(=\dfrac{4}{3 d}\)
\(d\) \(=4.0 \ \text{metres}\)

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 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 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 2017 HSC 13a

Using the cosine rule, find the value of `x` in the following diagram.  (3 marks)
 

Show Answers Only

`x = 11`

Show Worked Solution

`text(C)text(osine Rule:)`

`c^2` `= a^2 + b^2 – 2ab cos C`
`13^2` `= (x – 4)^2 + (x + 4)^2 – 2 (x – 4) (x + 4) cos 60^@`
`169` `= x^2 – 8x + 16 + x^2 + 8x + 16 – (x^2 – 16)`
`169` `= x^2 + 48`
`x^2` `= 121`
`:. x` `= 11, qquad (x != –11)`

Trigonometry, 2ADV T1 2016 HSC 12c

Square tiles of side length 20 cm are being used to tile a bathroom.

The tiler needs to drill a hole in one of the tiles at a point `P` which is 8 cm from one corner and 15 cm from an adjacent corner.

To locate the point `P` the tiler needs to know the size of the angle `theta` shown in the diagram.
 

 hsc-2016-12c

 
Find the size of the angle `theta` to the nearest degree.  (3 marks)

Show Answers Only

`69°\ text{(nearest degree)}`

Show Worked Solution

`α + theta = 90`

`text(Using the cosine rule,)`

`cos alpha` `= (20^2 + 15^2 – 8^2)/(2 xx 20 xx 15)`
  `= 0.935`
`alpha` `= 20.7…°`

 

`:. theta` `= 90 – 20.7…`
  `= 69.22…`
  `= 69°\ text{(nearest degree)}`

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 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 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)

Show Answers Only
  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 2006 HSC 1d

2006 1d

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

Show Answers Only

`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 2011 HSC 8a

In the diagram, the shop at  `S`  is 20 kilometres across the bay from the post office at  `P`. The distance from the shop to the lighthouse at  `L`  is 22 kilometres and  `/_ SPL`  is 60°.

Let the distance  `PL`  be  `x`  kilometres.
 

2011 8a
 

  1. Use the cosine rule to show that  `x^2-20x- 84 = 0`.    (1 mark)

  2. Hence, find the distance from the post office to the lighthouse. Give your answer correct to the nearest kilometre.    (2 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `24\ text(km)`
Show Worked Solution

i.  `text(Using the cosine rule)`

`cos 60^@` `= (x^2 + SP^2 – SL^2)/( 2 xx x xx 20)`
`1/2` `= (x^2 + 20^2 – 22^2)/(40x)`
`20x` `= x^2 – 84`

  
`:. x^2 – 20x – 84= 0\ \ \ text(… as required)`

 

ii.  `text(Find)\ \ LP:`

`x^2 – 20x – 84 = 0`

`x` `= (-b +- sqrt(b^2 – 4ac))/(2a)`
  `= (20 +- sqrt(20^2 – 4 xx 1 xx (–84)))/2`
  `= (20 +- sqrt(736))/2`
  `= 23.546…\ \ \ \ (x>0)`
  `= 24\ text(km)\ \ text{(nearest km)}`

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)`

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` `= 16-48= 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(.)`

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`