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Measurement, STD2 M6 2025 HSC 35

The triangle \(PTA\) is shown. The length of \(PA\) is 75 m and the length of \(PT\) is 51 m.

The angle of depression from \(T\) to \(A\) is 36°, and the angle \(PTA\) is obtuse.
 

Find the length of \(TA\). Give your answer correct to 2 decimal places.   (3 marks)

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\(TA=35.03 \ \text{m}\)

Show Worked Solution

\(\angle TAP=36^{\circ} \ \text {(alternate)}\)

\(\text{Using sine rule in} \ \triangle TAP:\)

\(\dfrac{\sin \angle PTA}{75}\) \(=\dfrac{\sin 36^{\circ}}{51}\)
\(\sin \angle PTA\) \(=75 \times \dfrac{\sin 36^{\circ}}{54}=0.864 \ldots\)
\(\angle PTA\) \(=\sin ^{-1}(0.864 \ldots)=180-59.81=120.19^{\circ}\ \ \text{(obtuse)}\)

 
\(\angle PTX=120.19-54=66.19^{\circ}\)

\(\angle TPA=90-66.19=23.81^{\circ}\ \left(180^{\circ}\ \text{in}\ \triangle \right)\)
 

\(\text{Using sine rule in} \ \triangle TAP:\)

\(\dfrac{TA}{\sin 23.81^{\circ}}\) \(=\dfrac{51}{\sin 36^{\circ}}\)
\(TA\) \(=\dfrac{51 \times \sin 23.81^{\circ}}{\sin 36^{\circ}}\)
  \(=35.03 \ \text{m (2 d.p.)}\)

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 2023 HSC 35

The diagram shows triangle `ABC`.
 

Calculate the area of the triangle, to the nearest square metre.  (3 marks)

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`147\ text{m}^2`

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`text{Using the sine rule:}`

`(CB)/sin60^@` `=12/sin25^@`  
`CB` `=sin60^@ xx 12/sin25^@`  
  `=24.590…`  

 
`angleACB=180-(60+25)=95^@\ \ text{(180° in Δ)}`
 

`text{Using the sine rule (Area):}`

`A` `=1/2 xx AC xx CB xx sin angleACB`  
  `=1/2 xx 12 xx 24.59 xx sin95^@`  
  `=146.98…`  
  `=147\ text{m}^2`  

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^{′}`

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`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 2021 HSC 37

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)

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`133°`

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`text(Using the sine rule:)`

♦♦ Mean mark 31%.
`sin theta/25` `= (sin 28°)/16`
`sin theta` `= (25 xx sin 28°)/16`
`sin theta` `= 0.73355`
`theta` `= 47°`
 
`:. angleABC` `= 180-47`
  `= 133°`

Measurement, STD2 M6 SM-Bank 1

The diagram shows a triangle with side lengths 25 cm and 47 cm and angle 30° and `theta`.
 

Find  `theta`  given  it is an obtuse angle. Give your answer to the nearest minute.  (3 marks)

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`109°57′`

Show Worked Solution

`text(Using sine rule:)`

`(sintheta)/47` `= (sin30°)/25`
`sintheta` `=47 xx (1/2)/25`
`sintheta` `= 47/50`
`theta` `= sin^(−1)\ 47/50`
  `= 70.05\ text(or)\ 109.94`
  `= 109.948…\ \ (theta\ text(is obtuse))`
  `= 109°57′`

Measurement, STD2 M6 2005 HSC 5 MC

Which formula should be used to calculate the distance between Toby and Frankie?

  1. `a/(sin A) = b/(sin B)`
  2. `c^2 = a^2 + b^2`
  3. `A = 1/2 ab\ sinC`
  4. `c^2 = a^2 + b^2 − 2ab\ cosC`
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`A`

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`text(The triangle is not a right-angled triangle,)`

`:.\ text(Not)\ B`

`text(Given the information on the diagram provides)`

`text(2 angles and 1 side, the sine rule will work best.)`

`a/sinA = b/sinB`

`=> A`

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 26d

Find the area of triangle `ABC`, correct to the nearest square metre.   (3 marks)

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`717\ text(m²)`    `text{(nearest m²)}`

Show Worked Solution
♦♦ Mean mark 32%.
TIP: The allocation of 3 marks to this question should flag the need for more than 1 step.
`cos/_C` `=(AC^2 + CB^2-AB^2)/(2 xx AC xx CB)`
  `=(50^2 + 40^2-83^2)/(2 xx 50 xx 40)`
  `= -0.69725…`
`/_C` `=134.2067…^@`

 

`text(Using Area) = 1/2 ab\ sinC :`
`text(Area)\ Delta ABC` `=1/2 xx 50 xx 40 xx sin134.2067…^@`
  `=716.828…`
  `=717\ text(m²)\ \ \ \ text{(nearest m²)}`

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 2010 HSC 9 MC

Three towns `P`, `Q`  and `R` are marked on the diagram.

The distance from `R` to `P` is 76 km.  `angle RQP=26^circ`  and  `angle RPQ=46^@.`
 

 

  What is the distance from  `P`  to  `Q`  to the nearest kilometre?

  1. `100\ text(km)`
  2. `125\ text(km)`
  3. `165\ text(km)`
  4. `182\ text(km)`
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`C`

Show Worked Solution
`angle QRP` `=180-(26+46)     (180^circ\ text(in) \ Delta)`
  `=108^circ`

 

`text{Using sine rule}`

`(PQ)/sin108^circ` `=76/sin26^circ`
`PQ` `=(76xxsin108^circ)/sin26^circ`
  `=164.88\ text(km)`

`=>  C`

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`