SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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 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
  1. `72^@ \ text{(nearest degree)}`
  2. `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 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 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
  1. `135°`
  2. `185°`
  3. `text{73° (nearest degree)}`
Show Worked Solution
i.   

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

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

`:. /_ PAQ` `= 90^@ + 45^@`
  `= 135^@`

 

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

`= 90^@ + 45^@ + 50^@`

`= 185^@\ \ \ text{(or  S 5°W)}`

 

iii.  `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 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
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `420\ text(m)\ text{(nearest m)}`
  3. `2.8\ text(ha)\ text{(2 sig. figures)}`
Show Worked Solution
i.   

 

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

`/_AOD = 360\ – 320 = 40^@`

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

 

ii.  `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.
 

iii.  `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 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(m²)`

Show Worked Solution
♦ Mean mark 50%
`/_` `AOB` `=21^@ + (360` `-310)`
  `=71^@`

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

`text(Area)` `=1/2` `xx 60` `xx 75` `xx sin71^@`
  `=2127.4167…`
  `=2127` `text(m²)\ \ \ text{(nearest m²)}`