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Measurement, STD1 M3 2024 HSC 7 MC

Consider the diagram shown.
 

Which of the following is the correct expression for the length of \(x\) ?

  1. \(20\, \cos 40^{\circ}\)
  2. \(20\, \sin 40^{\circ}\)
  3. \(\dfrac{20}{\cos 40^{\circ}}\)
  4. \(\dfrac{20}{\sin 40^{\circ}}\)
Show Answers Only

\(A\)

Show Worked Solution
\(\cos 40^{\circ}\) \(=\dfrac{x}{20}\)  
\(x\) \(=20\, \times \cos 40^{\circ}\)  

 
\(\Rightarrow A\)

Measurement, STD1 M3 2023 HSC 29

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. \(259^{\circ}\)
Show Worked Solution

a.    \(\text{In}\ \Delta XPY:\)

\(\angle PXY=180-120=60^{\circ}\)

\(\cos 60^{\circ}\) \(=\dfrac{XP}{15}\)  
\(XP\) \(=15\times \cos 60^{\circ}\)  
  \(=7.5\ \text{km}\)  

♦♦ Mean mark (a) 24%.

b.    \(\text{In}\ \Delta XPC:\)

\(\text{Let}\ \theta = \angle CXP\)

\(\cos \theta\) \(=\dfrac{7.5}{40}\)  
\(\theta\) \(=\cos^{-1} \Big(\dfrac{7.5}{40}\Big)\)  
  \(=79.193…\)  
  \(=79^{\circ}\ \text{(nearest degree)}\)  

 

\(\text{Bearing}\ C\ \text{from}\ X\) \(=180+79\)  
  \(=259^{\circ}\)  

♦♦♦♦ Mean mark (b) 9%.

Measurement, STD1 M3 2023 HSC 16

From the top of a vertical cliff 120 metres high, a boat is observed. The angle of depression of the boat from the top of the cliff is 18°, as shown in the diagram.
 

Find the distance of the boat from the base of the cliff. Give your answer to the nearest metre.  (2 marks)

Show Answers Only

\(369 \text{ m (nearest metre)}\)

Show Worked Solution

\(\text{Let }x\text{ be the distance from the cliff to the boat.}\)

\(\text{The angle at the boat}=18^{\circ}\text{ as it is alternate to the angle of depression from the cliff to the boat.}\)

\(\tan\theta\) \(=\dfrac{\text{opp}}{\text{adj}}\)
\(\tan 18^{\circ}\) \(=\dfrac{120}{x}\)
\(x\) \(=\dfrac{120}{\tan 18^{\circ}}\)
  \(=369.322\ldots \text{m}\)
  \(\approx 369 \text{ m (nearest metre)}\)

♦♦ Mean mark 30%.

Measurement, STD1 M3 2022 HSC 32

The diagram shows two right-angled triangles, `A B C` and `A B D`,

where `A C=35 \ text{cm}`, `B D=93 \ text{cm}`, `/_ A C B=41^@` and ` /_ A D B=\theta`.
 


 

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

Show Answers Only

`19^@6`′

Show Worked Solution

`text{In}\ Delta ABC:`

`tan 41^@` `=(AB)/35`  
`AB` `=35xxtan 41^@`  
  `=30.425…`  

  
`text{In}\ Delta ABD:`

`sin theta` `=(AB)/(BD)`  
  `=(30.425…)/93`  
`:.theta` `=sin^(-1)((30.425…)/93)`  
  `=19.09…`  
  `=19^@6`′`\ \ text{(nearest minute)}`  

♦♦♦ Mean mark 17%.

Measurement, STD1 M3 2021 HSC 28

A right-angled triangle `XYZ` is shown. The length of `XZ` is 16 cm and `angleYXZ = 30°`.
 


 

  1. Find the side length, `XY`, of the triangle in centimetres, correct to two decimal places.  (2 marks)

  2. Hence, find the area of triangle `XYZ` in square centimetres, correct to one decimal place.  (3 marks)

Show Answers Only
  1. `13.86\ text(cm)`
  2. `55.4\ text(cm)^2`
Show Worked Solution
♦♦ Mean mark part (a) 24%.
a.    `cos 30^@` `=(XY)/16`
  `XY` `=16 xx cos30^@`
    `=13.856…`
    `=13.86\ text(cm)\ \ text{(2 d.p.)}`

 

b.  `text(Using the sine rule:)`

♦♦♦ Mean mark part (b) 14%.
  `text(Area)\ DeltaXYZ` `= 1/2 ab sinC`
    `= 1/2 xx 16 xx 13.86 xx sin30^@`
    `=55.44`
    `=55.4\ text(cm)^2\ \ text{(1 d.p.)}`

Measurement, STD1 M3 2020 HSC 11

Consider the triangle shown.
 


 

  1. Find the value of `theta`, correct to the nearest degree.   (2 marks)

  2. Find the value of `x`, correct to one decimal place.   (2 marks)

Show Answers Only
  1. `39^@`
  2. `12.1 \ text{(to 1 d.p.)}`
Show Worked Solution
a.      `tan theta` `= frac{8}{10}`
  `theta` `= tan ^(-1) frac{8}{10}`
    `= 38.659…`
    `= 39^@ \ text{(nearest degree)}`

♦ Mean mark 44% and 39% for part (a) and (b) respectively.

 

b.     `text{Using Pythagoras:}`

`x` `= sqrt{8^2 + 10^2}`
  `= 12.806…`
  `= 12.8 \ \ text{(to 1 d.p.)}`

Measurement, STD1 M3 2019 HSC 31

Two right-angled triangles, `ABC` and `ADC`, are shown.
 


 

Calculate the size of angle `theta`, correct to the nearest minute.  (3 marks)

Show Answers Only

`41°4′\ \ text{(nearest minute)}`

Show Worked Solution

`text(Using Pythagoras in)\ DeltaACD:`

♦♦♦ Mean mark 15%.

`AC^2` `= 2.5^2 + 6^2`
  `= 42.25`
`:.AC` `= 6.5\ text(cm)`

 
`text(In)\ DeltaABC:`

`costheta` `= 4.9/6.5`
`theta` `= cos^(−1)\ 4.9/6.5`
  `= 41.075…`
  `= 41°4′31″`
  `= 41°5′\ \ text{(nearest minute)}`

Measurement, STD1 M3 2019 HSC 12

A surfer is 150 metres out to sea. From that point, the angle of elevation to the top of the cliff is 12°.
 


 

How high is the cliff, to the nearest metre?  (2 marks)

Show Answers Only

`32\ text(metres  (nearest m))`

Show Worked Solution

`tan12° = h/150`

♦♦ Mean mark 33%.

`h` `= 150 xx tan12°`
  `= 31.88…`
  `= 32\ text(metres  (nearest m))`

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 2017 HSC 26d

A sewer pipe needs to be placed into the ground so that it has a 2° angle of depression. The length of the pipe is 15 000 mm.
 


 

How much deeper should one end of the pipe be compared to the other end? Answer to the nearest mm.  (2 marks)

Show Answers Only

`523\ text{mm  (nearest mm)}`

Show Worked Solution

`text(Let)\ \ x = text(depth needed)`

`sin 2^@` `= x/(15\ 000)`
`x` `= 15\ 000 xx sin 2^@`
  `= 523.49…`
  `= 523\ text{mm  (nearest mm)}`

Measurement, STD2 M6 2017 HSC 8 MC

The diagram shows a right-angled triangle.
 


 

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

  1. `70°16^{′}`
  2. ` 70°17^{′}`
  3. `70°27^{′}`
  4. `70°28^{′}`
Show Answers Only

`B`

Show Worked Solution
`tan theta` `= text(opp)/text(adj)`
  `= 5.3/1.9`
  `= 2.789…`
COMMENT: An angle that has over 30″ (seconds) is rounded up to the next minute (i.e. rounded up to 70°17′).

 

`:. theta` `= 70.277…^@`
  `=70°16^{′}39.8^{″}`
  `= 70^@17^{′}`

 
`=>B`

Measurement, STD2 M6 2015 HSC 9 MC

From the top of a cliff 67 metres above sea level, the angle of depression of a buoy is 42°.
  

 

How far is the buoy from the base of the cliff, to the nearest metre?

  1. `60\ text(m)`
  2. `74\ text(m)`
  3. `90\ text(m)`
  4. `100\ text(m)`
Show Answers Only

`B`

Show Worked Solution
♦ Mean mark 49%.
COMMENT: The angle of depression is a regularly examined concept. Make sure you know exactly what it refers to.

`text(Let)\ x\ text(= distance of buoy from cliff base)`

`tan\ 42^@` `= 67/x`
`x\ tan\ 42^@` `= 67`
`x` `= 67/(tan\ 42^@)`
  `= 74.41…\ text(m)`

`⇒ B`

Measurement, STD2 M6 2005 HSC 25b

2UG-2005-25b

  1. Use Pythagoras’ theorem to show that `ΔABC` is a right-angled triangle.  (1 mark)

  2. Calculate the size of `∠ABC` to the nearest minute.  (2 marks)

Show Answers Only
  1. `text(Proof)`
  2. `67°23^{′}`
Show Worked Solution

i.   `ΔABC\ text(is right-angled if)\ \ a^2 + b^2 = c^2`

`a^2 + b^2` `= 5^2 + 12^2`
  `= 169`
  `= 13^2`
  `= c^2…\ text(as required.)`

MARKER’S COMMENT: Know your calculator process for producing an angle in minutes/seconds. Note >30 “seconds” rounds up to the higher “minute”.

 
ii.  `sin ∠ABC = 12/13`

`:.∠ABC` `= 67.38…°`
  `=67°22^{′}48^{″}`
  `= 67°23^{′}\ \ \ text{(nearest minute)}`

Measurement, STD2 M6 2006 HSC 3 MC

The angle of depression of the base of the tree from the top of the building is 65°. The height of the building is 30 m.

How far away is the base of the tree from the building, correct to one decimal place?
 


 

  1. 12.7 m
  2. 14.0 m
  3. 33.1 m
  4. 64.3 m
Show Answers Only

`B`

Show Worked Solution
 

`text(Let)\ d =\ text(distance from base to tree)`

`tan25^@` `=d/30`  
`:.d` `=30 xx tan25^@`  
  `=13.98…\ text{m}`  

 
`=>  B`

Measurement, STD2 M6 2008 HSC 14 MC

Danni is flying a kite that is attached to a string of length 80 metres. The string makes an angle of 55° with the horizontal.

How high, to the nearest metre, is the kite above Danni’s hand?
 

VCAA 2008 14 mc
 

  1.    46 m
  2. 66 m
  3. 98 m
  4. 114 m
Show Answers Only

`B`

Show Worked Solution

2UG 2008 14MC ans

`sin 55^@` `= h/80`
`h` `= 80 xx sin 55^@`
  `= 65.532…\ text(m)`

`=>  B`

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)

Show Answers Only
  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 23a

The point `A` is 25 m from the base of a building. The angle of elevation from `A` to the top of the building is 38°.
 

  1. Show that the height of the building is approximately 19.5 m.   (1 mark)

  2. A car is parked 62 m from the base of the building.

     

    What is the angle of depression from the top of the building to the car?

     

    Give your answer to the nearest minute.   (2 marks)

Show Answers Only

i.    `text{Proof  (See Worked Solutions)}`

ii.   `17°28^{′}`

Show Worked Solution

i.  `text(Need to prove height (h) ) ~~ 19.5\ text(m)`

`tan 38^@` `= h/25`
`h` `= 25 xx tan38^@`
  `= 19.5321…`
  `~~ 19.5\ text(m)\ \ text(… as required.)`

 

ii.  

`text(Let)\ \ /_ \ text(Elevation (from car) ) = theta`

♦♦ Mean mark 33%
MARKER’S COMMENT: If >30 “seconds”, round to the higher “minute”.
`tan theta` `= h/62`
  `= 19.5/62`
  `= 0.3145…`
`:. theta` `= 17.459…`
  `= 17°27^{′}33^{″}..`
  `=17°28^{′}\ \ text{(nearest minute)}`

 

`:./_ \ text(Depression to car) =17°28^{′}\ \ text{(alternate to}\ theta text{)}`

Measurement, STD2 M6 2009 HSC 4 MC

Which is the correct expression for the value of `x` in this triangle? 
 

 

  1. `8/cos30°` 
  2. `8/sin30°` 
  3. `8 xx cos30°`  
  4. `8 xx sin30°` 
Show Answers Only

`A`

Show Worked Solution
`cos30^@`  `= 8/x`
`:.x` `= 8/cos30^@` 

 
`=>  A`

Measurement, STD2 M6 2012 HSC 27d

A disability ramp is to be constructed to replace steps, as shown in the diagram.

The angle of inclination for the ramp is to be 5°.   
  

Calculate the extra distance, `d`, that the ramp will extend beyond the bottom step.

Give your answer to the nearest centimetre.   (3 marks)

Show Answers Only

 `386\  text(cm)`

Show Worked Solution

`text(Let the horizontal part of the ramp) = x\ text(cm)`

♦♦ Mean mark 35%
MARKER’S COMMENT:  The better responses used a diagram of a simplified version of the ramp as per the Worked Solution.
`tan5^@` `= 39/x`
`x` `= 39/tan5^@`
  `= 445.772…`
   
`text(S)text(ince)\  \ x` `= 60 + d`
`d` `=445.772-60`
  `=385.772\  text(cm)`
  `=386\ text(cm)\ \ text{(nearest cm)}`

Measurement, STD2 M6 2011 HSC 9 MC

Two trees on level ground, 12 metres apart, are joined by a cable. It is attached 2 metres above the ground to one tree and 11 metres above the ground to the other.

What is the length of the cable between the two trees, correct to the nearest metre? 

  1.  `9\ text(m)`
  2. `12\ text(m)`
  3. `15\ text(m)`
  4. `16\ text(m)`
Show Answers Only

`C`

Show Worked Solution

`text(Using Pythagoras)`

`c^2` `=12^2+9^2`
  `=144+81`
  `=225`
`:.c` `=15,\ \ c>0`

 
`=>C`

Measurement, STD2 M6 2011 HSC 4 MC

The angle of depression from a kookaburra’s feet to a worm on the ground is 40°. The worm is 15 metres from a point on the ground directly below the kookaburra’s feet. 
 

 How high above the ground are the kookaburra's feet, correct to the nearest metre?

  1. 10 m
  2. 11 m
  3. 13 m
  4. 18 m
Show Answers Only

`C`

Show Worked Solution
`  /_ \ text{Elevation (worm)}` `= 40^@`    `text{(alternate angles)}`
`tan 40^@` `=h/15`
`:. h` `=15xxtan 40^@`
  `=12.58…\ text(m)`

`=>C`

Measurement, STD2 M6 2012 HSC 4 MC

 Which expression could be used to calculate the value of `x` in this triangle? 
 

 

  1. `29 xx cos40^@`  
  2. `29 xx cos 50^@`  
  3. `cos40^@/29`  
  4. `cos50^@/29`  
Show Answers Only

`A`

Show Worked Solution
♦ Mean mark 42%
`cos40^@` `= x/29`
`x` `= 29xxcos40^@`

 
`=>  A`