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Measurement, STD2 M6 2025 HSC 14 MC

Points \( M \) and \( P \) are the same distance from a third point \(R\).

The bearing of \( M \) from \( R \) is 017° and the bearing of \( P \) from \( R \) is 107°.

Which of the following best describes the bearing of \(P\) from \(M\)?

  1. Between 000° and 090°
  2. Exactly 090°
  3. Between 090° and 180°
  4. Exactly 180°
Show Answers Only

\(C\)

Show Worked Solution

\(\angle MRP = 107-17=90^{\circ}\)

\(\angle RMP = \angle MPR = 45^{\circ}\ \text{(equilateral triangle)}\)

\(\text{Bearing of \(P\) from \(M\)}\ = 180-28=152^{\circ}\)

\(\Rightarrow C\)

Measurement, STD2 M6 2023 HSC 27

The diagram shows the location of three places `X`, `Y` and `C`.

`Y` is on a bearing of 120° and 15 km from `X`.

`C` is 40 km from `X` and lies due west of `Y`.

`P` lies on the line joining `C` and `Y` and is due south of `X`.
  

  1. Find the distance from `X` to `P`.  (2 marks)

  2. What is the bearing of `C` from `X`, to the nearest degree?  (2 marks)

Show Answers Only
  1. `7.5\ text{km}`
  2. `259^@`
Show Worked Solution

a.    `text{In}\ ΔXPY:`

`anglePXY=180-120=60^@`

`cos 60^@` `=(XP)/15`  
`XP` `=15 xx cos 60^@`  
  `=7.5\ text{km}`  

 
b.    `text{In}\ ΔXPC:`

`text{Let}\ \ theta = angleCXP`

`cos theta` `=7.5/40`  
`theta` `=cos^{-1}(7.5/40)`  
  `=79.193…`  
  `=79^@\ \ text{(nearest degree)}`  

 

`text{Bearing}\ C\ text{from}\ X` `=180+79`  
  `=259^@`  

♦ Mean mark (b) 39%.

Measurement, STD2 M6 2022 HSC 33

The diagram shows an aeroplane that was flying towards an airport at  `A` on a bearing of `135^@ text{T}`. When it was at point `O`, 20 km away from the airport at  `A`, the flight course was changed. The aeroplane landed at an airport at `B` directly south of `O`. The distance from `O` to `B` is 50 km.
 


 

  1. Show that the distance between the airport at  `A` and the airport at `B` is 38.5 km, correct to 1 decimal place.  (2 marks)

  2. Use the sine rule to find the angle `O B A` to the nearest degree.  (2 marks)

  3. What is the bearing of the airport at `B` from the airport at  `A` ?  (1 mark)

Show Answers Only
  1. `text{Proof}`
  2. `22^@`
  3. `202^@text{T}`
Show Worked Solution

a.   `angle AOB=180-135=45^@`
 

`text{Using the cosine rule:}`

`AB^2` `=20^2+50^2-2xx20xx50xxcos45^@`  
  `=1485.786…`  
`:.AB` `=38.54…`  
  `=38.5\ \text{km (to 1 d.p.)}`  

♦ Mean mark part (a) 50%.

 

b.   `text{Let}\ \ theta=angleOBA`

`text{Using the sine rule:}`

`sintheta/20` `=sin45^@/38.5`  
`sintheta` `=(sin45^@xx20)/38.5`  
`:.theta` `=sin^(-1)((sin45^@xx20)/38.5)`  
  `=21.55…^@`  
  `=22^@\ \ text{(nearest degree)}`  

♦♦ Mean mark part (b) 39%.

 

c. 

`text{Bearing of}\ B\ text{from}\ \ A`

`=180 + 22`

`=202^@text{T}`


♦♦ Mean mark part (c) 25%.

Measurement, STD2 M6 2020 HSC 31

Mr Ali, Ms Brown and a group of students were camping at the site located at `P`. Mr Ali walked with some of the students on a bearing of 035° for 7 km to location `A`. Ms Brown, with the rest of the students, walked on a bearing of 100° for 9 km to location `B`.
 


 

  1. Show that the angle `APB` is 65°.  (1 mark)

  2. Find the distance `AB`.  (2 marks)

  3. Find the bearing of Ms Brown's group from Mr Ali's group. Give your answer correct to the nearest degree.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `8.76\ text{km  (to 2 d.p.)}`
  3. `146^@`
Show Worked Solution
a.    `angle APB` `= 100 – 35`
    `= 65^@`

 

b.   `text(Using cosine rule:)`

Mean mark 53%.
`AB^2` `= AP^2 + PB^2 – 2 xx AP xx PB cos 65^@`
  `= 49 + 81 – 2 xx 7 xx 9 cos 65^@`
  `= 76.750…`
`:.AB` `= 8.760…`
  `= 8.76\ text{km  (to 2 d.p.)}`

 
c.

 
`anglePAC = 35^@\ (text(alternate))`

♦♦ Mean mark 22%.

`text(Using cosine rule, find)\ anglePAB:`

`cos anglePAB` `= (7^2 + 8.76 – 9^2)/(2 xx 7 xx 8.76)`  
  `= 0.3647…`  
`:. angle PAB` `= 68.61…^@`  
  `= 69^@\ \ (text(nearest degree))`  

 

`:. text(Bearing of)\ B\ text(from)\ A\ (theta)` 

`= 180 – (69 – 35)`

`= 146^@`

Measurement, STD2 M6 SM-Bank 4

The diagram shows three checkpoints A, B and C. Checkpoint C is due east of Checkpoint A. The bearing of Checkpoint B from Checkpoint A is N22°E and the bearing of Checkpoint C from Checkpoint B is S68°E. The distance between Checkpoint A and Checkpoint B is 42 kilometres.
 


 

  1. Mark the given information on the diagram and explain why `angleABC` is 90°.  (2 marks)

  2. Find the distance, to the nearest kilometre, between Checkpoint A and Checkpoint C.  (2 marks)

  3. If a runner is travelling 12.6 km/h, how long does it take her to travel between Checkpoint A and Checkpoint B, in hours and minutes?  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `112\ text{km}`
  3. `3text(h 20 mins)`
Show Worked Solution
i.   
`angle ABC` `= 22 + 68`
  `= 90°`

 

ii.  `text(In)\ \ DeltaABC,`

`cosangleBAC` `= (AB)/(AC)`
`cos68°` `= 42/(AC)`
`AC` `= 42/(cos68°)`
  `= 112.11…`
  `= 112\ text{km  (nearest km)}`

 

iii.    `text(Travel time)` `= text(dist)/text(speed)`
    `= 42/12.6`
    `= 3.333…`
    `= 3text(h 20 mins)`

Measurement, STD2 M6 2011 HSC 24c

A ship sails 6 km from `A` to `B` on a bearing of 121°. It then sails 9 km to `C`.  The
size of angle `ABC` is 114°.
 

  1. What is the bearing of `C` from `B`?   (1 mark)

  2. Find the distance `AC`. Give your answer correct to the nearest kilometre.   (2 marks)

  3. What is the bearing of `A` from `C`? Give your answer correct to the nearest degree.   (3 marks)

Show Answers Only
  1.  `055^@`
  2.  `13\ text(km)`
  3.  `261^@`
Show Worked Solution
STRATEGY: Important: Draw North-South parallel lines through major points to make the angle calculations easier!
i.     2011 HSC 24c

 `text(Let point)\ D\ text(be due North of point)\ B`

`/_ABD=180-121\ text{(cointerior with}\ \ /_A text{)}\ =59^@`

`/_DBC=114-59=55^@`   

`:. text(Bearing of)\ \ C\ \ text(from)\ \ B\ \ text(is)\ 055^@`

 

ii.    `text(Using cosine rule:)`

`AC^2` `=AB^2+BC^2-2xxABxxBCxxcos/_ABC`
  `=6^2+9^2-2xx6xx9xxcos114^@`
  `=160.9275…`
`:.AC` `=12.685…\ \ \ text{(Noting}\ AC>0 text{)}`
  `=13\ text(km)\ text{(nearest km)}`

 

iii.    `text(Need to find)\ /_ACB\ \ \ text{(see diagram)}`

MARKER’S COMMENT: The best responses clearly showed what steps were taken with working on the diagram. Note that all North/South lines are parallel.
`cos/_ACB` `=(AC^2+BC^2-AB^2)/(2xxACxxBC)`
  `=((12.685…)^2+9^2-6^2)/(2xx(12.685..)xx9)`
  `=0.9018…`
`/_ACB` `=25.6^@\ text{(to 1 d.p.)}`

 

`text(From diagram,)`

`/_BCE=55^@\ text{(alternate angle,}\ DB\ text(||)\ CE text{)}`

`:.\ text(Bearing of)\ A\ text(from)\ C`

  `=180+55+25.6`
  `=260.6`
  `=261^@\ text{(nearest degree)}`

Measurement, STD2 M6 2018 HSC 7 MC

The diagram shows the positions of towns `A`, `B` and `C`.

Town `A` is due north of town `B` and `angleCAB = 34°`
  


 

What is the bearing of town `C` from town `A`?

  1. 034°
  2. 146°
  3. 214°
  4. 326°
Show Answers Only

`C`

Show Worked Solution

`text(Bearing of Town)\ C\ text(from Town)\ A:`
 

`text(Bearing)` `= 180 + 34`
  `= 214^@`

 
`=>C`

Measurement, STD2 M6 2017 HSC 30c

The diagram shows the location of three schools. School `A` is 5 km due north of school `B`, school `C` is 13 km from school `B` and `angleABC` is 135°.
 


 

  1. Calculate the shortest distance from school `A` to school `C`, to the nearest kilometre.  (2 marks)

  2. Determine the bearing of school `C` from school `A`, to the nearest degree.  (3 marks)

Show Answers Only
  1. `17\ text{km  (nearest km)}`
  2. `213^@`
Show Worked Solution

i.   `text(Using cosine rule:)`

`AC^2` `= AB^2 + BC^2 – 2 xx AB xx BC xx cos135^@`
  `= 5^2 + 13^2 – 2 xx 5 xx 13 xx cos135^@`
  `= 285.923…`
`:. AC` `= 16.909…`
  `= 17\ text{km  (nearest km)}`

 

ii.   

`text(Using sine rule, find)\ angleBAC:`

♦♦ Mean mark 31%.
`(sin angleBAC)/13` `= (sin 135^@)/17`
`sin angleBAC` `= (13 xx sin 135^@)/17`
  `= 0.5407…`
`angleBAC` `= 32.7^@`

 

`:. text(Bearing of)\ C\ text(from)\ A`

`= 180 + 32.7`

`= 212.7^@`

`= 213^@`

Measurement, STD2 M6 2016 HSC 25 MC

The diagram shows towns `A`, `B` and `C`. Town `B` is 40 km due north of town `A`. The distance from `B` to `C` is 18 km and the bearing of `C` from `A` is 025°. It is known that  `∠BCA`  is obtuse.
 

2ug-2016-hsc-25-mc

 
What is the bearing of `C` from `B`?

  1.    `070°`
  2.    `095°`
  3.    `110°`
  4.    `135°`
Show Answers Only

`=> D`

Show Worked Solution

`text(Using the sine rule,)`

♦ Mean mark 39%.
`(sin∠BCA)/40` `= (sin25^@)/18`
`sin angle BCA` `= (40 xx sin25^@)/18`
  `= 0.939…`
`angle BCA` `= 180 – 69.9quad(angleBCA > 90^@)`
  `= 110.1°`

 

`:. text(Bearing of)\ C\ text(from)\ B`

`= 110.1 + 25qquad(text(external angle of triangle))`

`= 135.1`

 
`=> D`

Measurement, STD2 M6 2005 HSC 27c

2UG-2005-27c
 

The bearing of `C` from `A` is 250° and the distance of `C` from `A` is 36 km.

  1. Explain why  `theta`  is 110°.   (1 mark)

  2. If  `B`  is 15 km due north of  `A`, calculate the distance of  `C`  from  `B`, correct to the nearest kilometre.   (3 marks)

Show Answers Only
  1. `110^@`
  2. `text{43 km (nearest km)}`
Show Worked Solution

i.  `text(There is 360° about point)\ A`

`:.theta + 250^@` `= 360^@`
`theta` `= 110^@`

 

ii.   
`a^2` `= b^2 + c^2 − 2ab\ cos\ A`
`CB^2` `= 36^2 + 15^2 − 2 xx 36 xx 15 xx cos\ 110^@`
  `= 1296 + 225 −(text(−369.38…))`
  `= 1890.38…`
`:.CB` `= 43.47…`
  `= 43\ text{km  (nearest km)}`

Measurement, STD2 M6 2007 HSC 26a

The diagram shows information about the locations of towns  `A`,  `B`  and  `Q`.
 

 
 

  1. It takes Elina 2 hours and 48 minutes to walk directly from Town `A` to Town `Q`.

     

    Calculate her walking speed correct to the nearest km/h.    (1 mark)

  2. Elina decides, instead, to walk to Town `B` from Town `A` and then to Town `Q`.

     

    Find the distance from Town `A` to Town `B`. Give your answer to the nearest km.   (2 marks)

  3. Calculate the bearing of Town `Q` from Town `B`.   (1 mark)

Show Answers Only
  1. `5\ text(km/hr)\ text{(nearest km/hr)}`
  2. `18\ text(km)\ text{(nearest km)}`
  3. `236^@`
Show Worked Solution

i.  `text(2 hrs 48 mins) = 168\ text(mins)`

`text(Speed)\ text{(} A\ text(to)\ Q text{)}` `= 15/168`
  `= 0.0892…\ text(km/min)`

 

`text(Speed)\ text{(in km/hr)}` `= 0.0892… xx 60`
  `= 5.357…\ text(km/hr)`
  `= 5\ text(km/hr)\ text{(nearest km/hr)}`

 

ii.  

`text(Using cosine rule)`

`AB^2` `= 15^2 + 10^2 – 2 xx 15 xx 10 xx cos 87^@`
  `= 309.299…`
`AB` `= 17.586…`
  `= 18\ text(km)\ text{(nearest km)}`

 

`:.\ text(The distance from Town)\ A\ text(to Town)\ B\ text(is 18 km.)`

 

iii  
`/_CAQ` `= 31^@\ \ \ text{(} text(straight angle at)\ A text{)}`
`/_AQD` `= 31^@\ \ \ text{(} text(alternate angle)\ AC\ text(||)\ DQ text{)}`
`/_DQB` `= 87 – 31 = 56^@`
`/_QBE` `= 56^@\ \ \ text{(} text(alternate angle)\ DQ \ text(||)\ BE text{)}`

 

`:.\ text(Bearing of)\ Q\ text(from)\ B`

`= 180 + 56`

`= 236^@`

Measurement, STD2 M6 2008 HSC 17 MC

The diagram shows the position of  `Q`,  `R`  and  `T`  relative to  `P`.
 

VCAA 2008 17 mc

 
In the diagram,

`Q`  is south-west of  `P`

`R`  is north-west  of  `P`

`/_QPT`  is 165°
 

What is the bearing of  `T`  from  `P`?

  1.    `060^@`
  2.    `075^@`
  3.    `105^@` 
  4.    `120^@`
Show Answers Only

`A`

Show Worked Solution

VCAA 2008 17 mci

`/_QPS=45^@\ \ \ text{(} Q\ text(is south west of)\ Ptext{)}`

`/_TPS = 165 – 45 = 120^@`

`:.\ /_NPT = 60^@\ \ text{(} 180^@\ text(in straight line) text{)}`

`=>  A`

Measurement, STD2 M6 2014 HSC 23 MC

The following information is given about the locations of three towns `X`, `Y` and `Z`: 

• `X` is due east of  `Z`

• `X` is on a bearing of  145°  from  `Y` 

• `Y` is on a bearing of  060°  from  `Z`. 

Which diagram best represents this information?
 

HSC 2014 23mci

Show Answers Only

`C`

Show Worked Solution
♦ Mean mark 38%
COMMENT: Drawing a parallel North/South line through `Y` makes this question much simpler to solve.

`text(S)text(ince)\ X\ text(is due east of)\ Z`

`=> text(Cannot be)\ B\ text(or)\ D`
 

 
`text(The diagram shows we can find)`

`/_ZYX = 60 + 35^@ = 95^@`

`text(Using alternate angles)\ (60^@)\ text(and)`

`text(the)\ 145^@\ text(bearing of)\ X\ text(from)\ Y`

`=>  C`

Measurement, STD2 M6 2009 HSC 27b

A yacht race follows the triangular course shown in the diagram. The course from  `P`  to  `Q`  is 1.8 km on a true bearing of 058°.

At  `Q`  the course changes direction. The course from  `Q`  to  `R`  is 2.7 km and  `/_PQR = 74^@`.
 

 2009-2UG-27b
 

  1. What is the bearing of  `R`  from  `Q`?   (1 mark)

  2. What is the distance from  `R`  to  `P`?     (2 marks)

  3. The area inside this triangular course is set as a ‘no-go’ zone for other boats while the race is on.

     

    What is the area of this ‘no-go’ zone?   (1 mark)

Show Answers Only
  1. `312^@`
  2. `2.8\ text(km)\ \ \ text{(1 d.p.)}`
  3. `2.3\ text(km²)\ \ \ text{(1 d.p.)}`
Show Worked Solution
i.    2UG-2009-27b-Answer

`/_ PQS = 58^@ \ \ \ (text(alternate to)\ /_TPQ)`

♦♦♦ Mean mark 18%.
TIP: Draw North-South parallel lines through relevant points to help calculate angles as shown in the Worked Solutions.

`text(Bearing of)\ R\ text(from)\ Q`

`= 180^@ + 58^@ + 74^@`
`= 312^@`

 

ii.   `text(Using Cosine rule:)`

♦ Mean mark 36%
`RP^2` `=RQ^2` + `PQ^2` `- 2` `xx RQ` `xx PQ` `xx cos` `/_RQP`
  `= 2.7^2` + `1.8^2` `- 2` `xx 2.7` `xx 1.8` `xx cos74^@`
  `=7.29 + 3.24\ – 2.679…`
  `=7.851…`
`:.RP` `= sqrt(7.851…)`
  `=2.8019…`
  `~~ 2.8\ text(km)  (text(1 d.p.) )`

 

iii.   `text(Using)\ \ A = 1/2 ab sinC`

♦ Mean mark 44%
`A` `= 1/2` `xx 2.7` `xx 1.8` `xx sin74^@`
  `= 2.3358…`
  `= 2.3\ text(km²)`

 

`:.\ text(No-go zone is 2.3 km²)`

Measurement, STD2 M6 2010 HSC 10 MC

A plane flies on a bearing of  150° from  `A`  to  `B`.
 

Capture3

 
What is the bearing of  `A` from `B`?

  1. `30^@`
  2. `150^@`
  3. `210^@`
  4. `330^@`
Show Answers Only

`D`

Show Worked Solution
♦♦ Mean mark 34%

Capture3-i
 

`/_TBA=30^@\ \ \ text{(angle sum of triangle)}`

`:.\ text(Bearing of)\ A\ text{from}\ B`

`=360-30`

`=330^@`

`=>  D`

Measurement, STD2 M6 2012 HSC 20 MC

Town `B` is 80 km due north of Town `A` and 59 km from Town `C`.

Town `A` is 31 km from Town `C`.
 

2012 20 mc
 

 What is the bearing of Town `C` from Town `B`?  

  1.   `019^@`
  2.   `122^@` 
  3.   `161^@` 
  4.   `341^@` 
Show Answers Only

`C`

Show Worked Solution
♦ Mean mark 49%

`text(Using the cosine rule:)`

`cos\ /_B` `= (a^2 + c^2 -b^2)/(2ac)`
  `= (59^2 + 80^2 -31^2)/(2 xx 59 xx 80)`
  `= 0.9449…`
`/_B` `= 19^@\  text((nearest degree))`

 

`:.\ text(Bearing of Town C from B) = 180-19= 161^@`

`=>  C`