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Measurement, STD2 M6 2019 HSC 12 MC

An owl is 7 metres above ground level, in a tree. The owl sees a mouse on the ground at an angle of depression of 32°.

How far must the owl fly in a straight line to catch the mouse, assuming the mouse does not move?

  1.  3.7 m
  2.  5.9 m
  3.  8.3 m
  4.  13.2 m
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`D`

Show Worked Solution

`text(Let)\ \ OM = text(Flight distance)`

♦ Mean mark 36%.

`sin32°` `= 7/(OM)`
`:. OM` `= 7/(sin32°)`
  `= 13.2\ text(m)`

 
`=> D`

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)

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`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 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)`
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`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 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
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`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 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 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)

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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 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
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`C`

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

`=>C`