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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}\)

\(\Rightarrow C\)

♦ Mean mark 41%.

Measurement, STD2 M6 2024 HSC 40

A compass radial survey is shown.
 

Given that \(AC\) is a straight line, find the bearing of \(C\) from \(O\), correct to the nearest degree.   (3 marks)

Show Answers Only

\(\text{Bearing of}\ C\ \text{from}\ O =220^{\circ}\)

Show Worked Solution

\(\text{In}\ \Delta ODA,\ \text{using cosine rule:}\)

\(\cos \angle DOA\) \(=\dfrac{38^2+42^2-67.6^2}{2 \times 38 \times 42}\)  
  \(=-0.4223…\)  
\(\angle DOA\) \(=115.0^{\circ}\)  

 
\(\text{Let}\ X\ \text{be a point directly north of point}\ O:\)

\(\angle DOX=360-285=75^{\circ}\)

\(\angle XOA=115-75=40^{\circ}\)

\(\angle AOB=110-40=70^{\circ}\)

\(\angle COB=180-70=110^{\circ}\ \ (AOC\ \text{is a straight line)}\)

\(\therefore\ \text{Bearing of}\ C\ \text{from}\ O = 110+110=220^{\circ}\)

♦♦♦ Mean mark 19%
STRATEGY:  The information in \(\Delta ODA\) strongly suggests finding \(\angle DOA\) is a good strategy to explore.

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

a.    `7.5\ text{km}`

b.    `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

a.    `text{Proof}`

b.    `22^@`

c.    `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 2021 HSC 39

The diagram shows a compass radial survey of the field `ABCD`.
 

  1. Triangle `COB` has an area of 466 m².
  2. Find the size of acute angle `COB`, correct to the nearest degree.   (2 marks)

  3. A farmer wants to put a fence around the triangle `DOC`.
  4. Find the length of fencing required. Give your answer in metres correct to one decimal place.   (3 marks)

Show Answers Only

a.    `72^@ \ text{(nearest degree)}`

b.    `106.8 \ text{m (1 d.p.)}`

Show Worked Solution
a.    `text{Using sine rule:}`

♦ Mean mark part (a) 34%.
`text{Area}` `= 1/2 ab sinC`  
`466` `= 1/2 xx 28 xx 35 xx sin∠ COB`  
`sin ∠COB` `= 466/490`  
`∠COB` `= sin^{-1} (466/490)`  
  `= 71.99`  
  `= 72^@ \ text{(nearest degree)}`  

 

b.

♦♦ Mean mark part (b) 33%.

   

`∠DOC` `= 360-(30 + 80 + 70 + 72)`  
  `= 108^@`  

 

`text{Using the cosine rule:}`

`DC^2` `=31^2 + 28^2-2 xx 31 xx 28 xx cos 108^@`  
  `= 2281.4535`  
`DC` `= 47.76 … \ text{m}`  
  
 
`:. \ text{Length of fence}` `= 47.8 + 31 + 28`  
  `= 106.8 \ text{m (1 d.p.)}`  

Measurement, STD2 M6 2021 HSC 14 MC

Consider the diagram below.
 


 

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

  1. `025^@`
  2. `065^@`
  3. `115^@`
  4. `295^@`
Show Answers Only

`D`

Show Worked Solution

♦♦ Mean mark 28%.

`\text{Bearing (A from B)}= 270 + 25= 295^@`

    
`=> D`

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

a.    `text(See Worked Solutions)`

b.    `8.76\ text{km  (to 2 d.p.)}`

c.    `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 2019 HSC 35

A compass radial survey shows the position of four towns  `A`, `B`, `C` and `D`  relative to point `O`.
 


 

The area of triangle  `BOC`  is 198 km².

Calculate the bearing of town  `C`  from point  `O`, correct to the nearest degree.   (3 marks)

Show Answers Only

`207^@`

Show Worked Solution

`text(In)\ DeltaCOB,`

♦♦ Mean mark 26%.

`text(Area) = 1/2 ab sinc`

`198` `= 1/2 xx 16 xx 25 xx sinangleCOB`
`sinangleCOB` `= (2 xx 198)/(16 xx 25)`
  `= 0.99`
`angleCOB` `= sin^(−1)0.99`
  `= 81^@53^{′}`

 

`:. text(Bearing of)\ C\ text(from)\ O` `= 125^@ + 81^@53^{′}= 206^@53^{′}`
  `= 207^@\ \ (text(nearest degree))`

Measurement, STD2 M6 2019 HSC 4 MC

Which compass bearing is the same as a true bearing of 110°?

  1. `text(S20°E)`
  2. `text(S20°W)`
  3. `text(S70°E)`
  4. `text(S70°W)`
Show Answers Only

`C`

Show Worked Solution

♦ Mean mark 47%.

`110° = text(S70°E)`

`=> C`

Measurement, STD2 M6 EQ-Bank 25

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

a.    `text(See Worked Solutions)`

b.    `112\ text{km}`

c.    `3\ text(h 20 mins)`

Show Worked Solution
a.    

`angle ABC= 22 + 68= 90^@`
  

b.    `text(In)\ \ DeltaABC,`

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

 

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

Measurement, STD2 M6 EQ-Bank 7 MC

Jeet walks 5 km from his home on a bearing of 153°. He then walks due north until he arrives a point which is due east of his home.

How far east, to the nearest 0.1 km, is Jeet from home?

  1. 2.3 km
  2. 2.5 km
  3. 4.9 km
  4. 9.8 km
Show Answers Only

`A`

Show Worked Solution

`text(Jeet finishes at)\ P`

`text(Find)\ \ OP:`

`cos 63°` `= (OP)/5`
`:. OP` `= 5 xx cos 63°= 2.26995…`

 
`=> A`

Measurement, STD2 M6 EQ-Bank 5 MC

Which of the following expresses S65°W as a true bearing?

  1. 065°
  2. 155°
  3. 245°
  4. 295°
Show Answers Only

`C`

Show Worked Solution

`text(True bearing)= 180 + 65= 245°`
  

`=> C`

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

a.    `055^@`

b.    `13\ text(km)`

c.     `261^@`

Show Worked Solution
STRATEGY: Important: Draw North-South parallel lines through major points to make the angle calculations easier!
a.        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^@`

 

b.    `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)}`

 

c.    `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` `=cos^(-1)0.9018…`
  `=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

i.    `17\ text{km  (nearest km)}`

ii.   `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 2015 HSC 7 MC

The diagram shows a radial survey of a field `ABCD`.
 

2UG 2015 7MC
 

In triangle `AOB`, what is the size of `∠AOB` ?

  1. `51^@`
  2. `111^@`
  3. `125^@`
  4. `249^@`
Show Answers Only

`B`

Show Worked Solution

2UG 2015 7MC Answer

`text(Let)\ P\ text(be due North of)\ O`
  

`∠POB= 51^@`

`∠POA= 360-300= 60^@`
  

`:.∠AOB= 51 + 60= 111^@`
  

`=>B`

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

a.    `110^@`

b.    `text{43 km (nearest km)}`

Show Worked Solution

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

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

 

b.    
`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-(-369.38…)`
  `= 1890.38…`
`:.CB` `= 43.47…`
  `= 43\ text{km  (nearest km)}`

Measurement, STD2 M6 2004 HSC 24b

The diagram shows a radial survey of a piece of land.
 


 

  1. `Q` is south-east of `A`. What is the size of angle `PAQ`?   (1 mark)

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

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

Show Answers Only

a.    `135°`

b.    `185°`

c.    `text{73° (nearest degree)}`

Show Worked Solution
a.    

`text(Let)\ E\ text(be directly east of)\ A`

`/_ EAQ = 45^@\ \ \ \ text{(given)}`

`:. /_ PAQ= 90^@ + 45^@= 135^@`
  

b.    `text(Bearing of)\ R\ text(from)\ A`

`= 90^@ + 45^@ + 50^@= 185^@\ \ \ text{(or  S 5°W)}`
  

c.    `text(Using cosine rule in)\ Delta PAB`

` cos\ /_ PAB` `= (31^2 + 28^2-35^2) / (2 xx 31 xx 28)`
  `= 0.2995…`
`:. /_ PAB` `= 72.570…^@`
  `= 73^@\ \ \ \ text{(nearest degree)}`

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

a.    `5\ text(km/hr)\ text{(nearest km/hr)}`

b.    `18\ text(km)\ text{(nearest km)}`

c.    `236^@`

Show Worked Solution

a.    `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)}`

 

b.    

`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.)`
  

c.    
`/_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 28b

A radial compass survey of a sports centre is shown in the diagram. 
 

 
 

  1. Show that the size of angle  `AOB`  is  114°.   (1 mark)

  2. Calculate the length of the boundary `AB`, to the nearest metre.   (2 marks)

  3. Find the area of triangle `AOB` in hectares, correct to two significant figures.   (3 marks)

Show Answers Only

a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `420\ text(m)\ text{(nearest m)}`

c.    `2.8\ text(ha)\ text{(2 sig. figures)}`

Show Worked Solution
a.    

  
`text(Let)\ D\ text(be directly north of)\ O`

`/_AOD = 360-320 = 40^@`

`:.\ /_AOB = 40 + 74 = 114^@\ \ \ text(… as required)`

 

b.    `text(Using cosine rule)`

`AB^2` `= AO^2 + BO^2-2 xx AO xx BO xx cos /_AOB`
  `= 287^2 + 211^2-2 xx 287 xx 211 xx cos 114^@`
  `= 126\ 890-121\ 114 (-0.4067…)`
  `= 176\ 151.50…`
`AB` `= 419.704…`
  `= 420\ text(m)\ text{(nearest m)}`
♦ Mean marks of 42% and 41% for parts (ii) and (iii) respectively.
 

c.    `text(Using)\ A = 1/2 ab sin C,`

`text(Area)\ Delta AOB` `= 1/2 xx 287 xx 211 xx sin 114^@`
  `= 27\ 660.786…\ text(m²)`
  `= 2.7660…\ text(ha)\ \ \ \ text{(1 ha = 10 000 m²)}`
  `= 2.8\ text(ha)\ text{(2 sig figures)}`

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

a.    `312^@`

b.    `2.8\ text(km)\ \ \ text{(1 d.p.)}`

c.    `2.3\ text(km²)\ \ \ text{(1 d.p.)}`

Show Worked Solution
a.     2UG-2009-27b-Answer

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

♦♦♦ Mean mark (a) 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^@`
  

b.    `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.) )`

 

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

♦ Mean mark (c) 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 2013 HSC 28a

A compass radial survey of the field `ABCD` has been conducted from `O`.
 

2013 28a
 

Find the area of the section `ABO`, to the nearest square metre.   (2 marks)

 

Show Answers Only

 `2127\ text(m)^2`

Show Worked Solution
♦ Mean mark 50%

`/_AOB=21 + (360-310)=71^@`
  

`text(Using Area) = 1/2 ab sinC`

`text(Area)` `=1/2xx 60xx 75xx sin71^@`
  `=2127.4167…=2127 text(m)^2\ \ (text(nearest m)^2)`

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`