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Measurement, STD2 M6 2024 HSC 36

The diagram shows two vertical flagpoles, \(BE\) and \(CD\), set on sloping ground.
 

  1. What is the height of the flagpole \(BE\), correct to 1 decimal place?   (2 marks)
  2. What is the height of the flagpole \(CD\), correct to 1 decimal place?   (2 marks)
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a.   \(BE=25.4\ \text{m}\)

b.   \(CD=19.7\ \text{m}\)

Show Worked Solution

a.   \(\text{Using the sine rule:}\)

\(\dfrac{BE}{\sin 27^{\circ}}\) \(=\dfrac{53.8}{\sin 106^{\circ}}\)  
\(\therefore BE\) \(=\dfrac{53.8 \times \sin 27^{\circ}}{\sin 106^{\circ}}\)  
  \(=25.408…\)  
  \(=25.4\ \text{m (1 d.p.)}\)  

 

b.   \(CD=EB-XB\)

\(\text{Consider}\ \Delta XBC:\)

\(\angle XBC=180-106=74^{\circ}, \ XC=ED=20\)

\(\tan 74^{\circ}\) \(=\dfrac{20}{XB}\)  
\(XB\) \(=\dfrac{20}{\tan 74^{\circ}}\)  
  \(=5.73\ \text{m}\)  

 
\(CD=25.4-5.73=19.7\ \text{m (1 d.p.)}\)

♦♦ Mean mark (b) 35%.

Measurement, STD2 M6 2022 HSC 26

The diagram shows two right-angled triangles, `ABC` and `ABD`,

where `AC=35 \ text{cm},BD=93 \ text{cm}, /_ACB=41^(@)` and `/_ADB=theta`.
 
     

Calculate the size of angle `theta`, to the nearest minute.  (4 marks)

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`19^@6^{′}`

Show Worked Solution

`text{In}\ Delta ABC:`

`cos 41^@` `=35/(BC)`  
`BC` `=35/(cos 41^@)`  
  `=46.375…`  

 
`angle BCD = 180-41=139^@`
 

`text{Using sine rule in}\ Delta BCD:`

`sin theta/(46.375)` `=sin139^@/93`  
`sin theta` `=(sin 139^@ xx 46.375)/93`  
`:.theta` `=sin^(-1)((sin 139^@ xx 46.375)/93)`  
  `=19.09…`  
  `=19^@6^{′}\ \ text{(nearest minute)}`  

♦ Mean mark 50%.

Measurement, STD2 M6 2018 HSC 30c

The diagram shows two triangles.

Triangle `ABC` is right-angled, with  `AB = 13 text(cm)`  and  `/_ABC = 62°`.

In triangle  `ACD, \ AD = x\ text(cm)`  and  `/_DAC = 40°`. The area of triangle  `ACD`  is 30 cm².
 

 
What is the value of `x`, correct to one decimal place?  (3 marks)

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`8.1\ text{cm  (1 d.p.)}`

Show Worked Solution

`text(Find)\ AC:`

♦ Mean mark 39%.

`sin62°` `= (AC)/13`
`AC` `= 13 xx sin62°`
  `= 11.478…`

 
`text(Using the sine rule in)\ DeltaACD :`

`text(Area)` `= 1/2 xx AC xx AD xx sin40°`
`30` `= 1/2 xx 11.478… xx x xx sin40°`
`:.x` `= (30 xx 2)/(11.478… xx sin40°)`
  `= 8.13…`
  `= 8.1\ text{cm  (1 d.p.)}`

Measurement, STD2 M6 2015 HSC 30e

From point `S`, which is 1.8 m above the ground, a pulley at `P` is used to lift a flat object `F`. The lengths `SP` and `PF` are 5.4 m and 2.1 m respectively. The angle `PSC` is 108°.
 

 

  1. Show that the length  `PC`  is 6.197 m, correct to 3 decimal places.  (1 mark)

  2. Calculate `h`, the height of the object above the ground.  (4 marks)

Show Answers Only
  1. `6.197\ text{m  (to 3 d.p.)  … as required}`
  2. `1.37\ text{m  (to 2 d.p.)}`
Show Worked Solution
i.   

2UG 2015 30e Answer

`text(Show)\ PC = 6.197\ text(m)`

♦ Mean mark 41%.

`text(Using the cosine rule in)\ Delta PSC` 

`PC^2` `= PS^2 + SC^2-2 xx PS xx SC xx cos\ 108^@`
  `= 5.4^2 + 1.8^2-2 xx 5.4 xx 1.8 xx cos\ 108^@`
  `= 38.4072…`
`:.PC` `= 6.19736…`
  `= 6.197\ text{m  (to 3 d.p.)  …as required}`

 

ii.   `text(Let)\ \ SD⊥PE`

♦♦ Mean mark below 19%.
STRATEGY: Finding `PC` in part (i) and needing `PE` to find `h` should flag the strategy of finding `EC` and using Pythagoras.

`∠DSC\ text(is a right angle)`

`:.∠DSP = 108^@-90^@ = 18^@`

 

`text(In)\ ΔPDS`

`cos\ 18^@` `= (DS)/5.4`
 `DS` `= 5.4 xx cos\ 18^@`
  `= 5.1357…\ text(m)`

 
`EC = DS = 5.1357…\ text{m  (opposite sides of rectangle}\ DECS text{)}`
 

`text(Using Pythagoras in)\ Delta PEC:`

`PE^2 + EC^2 = PC^2`

`PE^2 + 5.1357^2` `= 6.197^2`
`PE^2` `= 12.027…`
`PE` `= 3.468…\ text(m)`

 
`text(From the diagram,)`

`h` `= PE-PF`
  `= 3.468…-2.1`
  `= 1.368…`
  `= 1.37\ text{m  (to 2 d.p.)}`

Measurement, STD2 M6 2006 HSC 24b

A 130 cm long garden rake leans against a fence. The end of the rake is 44 cm from the base of the fence.

  1. If the fence is vertical, find the value of `theta` to the nearest degree.  (2 marks)
      
          2UG-2006-24b-i

     

  2. The fence develops a lean and the rake is now at an angle of 53° to the ground. Calculate the new distance (`x` cm) from the base of the fence to the head of the rake. Give your answer to the nearest centimetre.  (2 marks)
     
          2UG-2006-24b-ii

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  1. `text{70°  (nearest degree)}`
  2. `text{109 cm  (nearest cm)}`
Show Worked Solution
i.   

2UG-2006-24b1 Answer
`cos theta` `= 44/130`
  `= 70.216… ^@`
  `= 70^@\ \ \ text{(nearest degree)}`

 

ii.   

2UG-2006-24b2 Answer

`text(Using cosine rule)`

`x^2` `= 130^2 + 44^2-2 xx 130 xx 44 xx cos 53^@`
  `= 11\ 951.23…`
`x` `= 109.32…`
  `= 109\ text{cm  (nearest cm)}`

Measurement, STD2 M6 2007 HSC 25b

The angle of depression from `J` to `M` is 75°. The length of `JK` is 20 m and the length of `MK` is 18 m.
 

 
 

Copy or trace this diagram into your writing booklet and calculate the angle of elevation from `M` to `K`. Give your answer to the nearest degree.   (3 marks)

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`58^@`

Show Worked Solution

`/_AJL = 90^@`

`/_MJL = 90 – 75 = 15^@`

`text(Using sine rule in)\ Delta MJK`

`text(Let)\ /_JMK = x^@`

`20/sin x` `= 18/sin15^@`
`18 sin x` `= 20 xx sin 15^@`
`sin x` `= (20 xx sin 15^@)/18 = 0.2875…`
`x` `= 16.71…^@`
   
`/_JML = 75^@\ text{(} text(alternate angles,)\ ML \ text(||) \ AJ text{)}`

 

`:.\ /_KML` `= 75^@ – 16.71…`
  `= 58.287…^@`
  `= 58^@\ \ \ text{(nearest degree)}`

 

`:.\ text(Angle of Elevation from)\ M\ text(to)\ K\ text(is)\ 58^@.`

Measurement, STD2 M6 2010 HSC 24d

The base of a lighthouse, `D`, is at the top of a cliff 168 metres above sea level. The angle of depression from `D` to a boat at `C` is 28°. The boat heads towards the base of the cliff, `A`, and stops at `B`. The distance `AB` is 126 metres.
 

  1. What is the angle of depression from `D` to `B`, correct to the nearest degree?   (3 marks)

  2. How far did the boat travel from `C` to `B`, correct to the nearest metre?   (2 marks)

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  1. `53^circ`
  2. `190\ text(m)`
Show Worked Solution
♦♦ Mean mark 31%
i.    `tan/_ADB` `=126/168`
  ` /_ADB` `=36.8698…`
    `=36.9^circ\ \ \ \ text{(to 1 d.p)}` 

 

`/_text(Depression)\ D\ text(to)\ B` `=90-36.9`
  `=53.1`
  `=53^circ\ text{(nearest degree)}`

 

ii.     `text(Find)\ CB:`

♦♦ Mean mark 31%
MARKER’S COMMENT: Solve efficiently by using right-angled trigonometry. Many students used non-right angled trig, adding to the calculations and the difficulty.
`/_ADC+28` `=90`
 `/_ADC` `=62^circ`
`tan 62^circ` `=(AC)/168`
`AC` `=168xxtan 62^circ`
  `=315.962…`

 

`CB` `=AC-AB`
  `=315.962…-126`
  `=189.962…`
  `=190\ text(m (nearest m))`

Measurement, STD2 M6 2009 HSC 22 MC

In the diagram, `AD` and `DC` are equal to 30 cm. 
 

2UG-2009-22MC
 

 What is the length of `AB` to the nearest centimetre? 

  1.    `28\ text(cm)`
  2.    `31\ text(cm)` 
  3.    `34\ text(cm)`
  4.    `39\ text(cm)` 
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`A`

Show Worked Solution
♦ Mean mark of 35%

`Delta ADC\ text(is isosceles)`

`/_DAB = /_DCA` `= x^@`
`2x + 80^@` `= 180^@\ \ \ (text{Angle sum of}\ DeltaADC)`
`2x` `= 100^@`
`x` `= 50^@`

 

`/_ DBA` `= 180\-(50 + 60)\ \ \ (text{Angle sum of}\ Delta ADB)`
  `= 70^@`

 

`text(Using sine rule:)`

`(AB)/sin60` `= 30/sin70`
`AB` `= (30 xx sin60)/sin70`
  `= 27.648…\ text(cm)`

`=>  A`

Measurement, STD2 M6 2012 HSC 29c

Raj cycles around a course. The course starts at `E`, passes through `F`, `G` and `H` and finishes at `E`. The distances `EH` and `GH` are equal.
  

2012 29c

  1. What is the length of `EF`, to the nearest kilometre?  (2 marks)

  2. What is the total distance that Raj cycles, to the nearest kilometre?   (3 marks)

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  1. `22\ text{km    (nearest km)}`
  2. `202\ text{km    (nearest km)}`
Show Worked Solution

i.   `text(Find)\ EF:`

♦ Mean mark 48%.
`/_ FGE` `= 180\-(139 + 31) \ \ text{(angle sum of}\ Delta EFGtext{)}`
  `= 180-170`
  `= 10^@`

 
`text(Using Sine rule:)`

`(EF)/sin10^@` `= 82/sin139^@`
`EF` `= (82 xx sin10^@)/sin139^@`
  `= 21.70406…`
  `= 22\  text{km (nearest km)}`

 

ii.  `text(Let)\ \ d = text(total distance cycled)`

`text(Find)\ EH`

`text(S)text(ince)\ Delta EGH\ text(is isosceles, and)\ /_EHG = 90^@`

`/_GEH = /_HGE = 45^@`

`text{(angles opposite equal sides in}\ Delta EGHtext{)}`

♦ Mean mark 37%.
MARKER’S COMMENT: Students could also have used Pythagoras or the Sine rule to calculate `GH`.
`sin45^@` `= (GH)/82`
`GH` `= 82 xx sin45^@`
  `= 57.983…`

 

`:. d` `= EF + FG + GH + EH`
  `= 21.704… + 64 + 57.983 + 57.983`
  `= 201.66…`
  `= 202 \ text{km (nearest km)}`

Measurement, STD2 M6 2013 HSC 24 MC

What is the value of  `theta`,  to the nearest degree?

2013 24 mc

  1.    `21^@`
  2.    `32^@`
  3.    `43^@`
  4.    `55^@`
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`C`

Show Worked Solution
`a/sinA` `=b/sinB`
`82/sinA` `=100/sin26`
`sin A` `=(82 xx sin26)/100`
  `=0.35946…`
`/_A` `=21^@\ \ \ \ text{(nearest degree)}`

 
`text(S)text(ince)\   180^@\ text(in)\ Delta:`

`90+26+(theta+21)` `=180`
`theta` `=43^@`

 
`=>  C`