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Measurement, STD2 M6 2011 FUR1 2


 

The point `Q` on building `B` is visible from the point `P` on building `A`, as shown in the diagram above.

Building `A` is 16 metres taller than building `B`.

The horizontal distance between point `P` and point `Q` is 23 metres.

Calculate the angle of depression of point `Q` from point `P`. Give your answer correct to 1 decimal place.   (2 marks)

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`34.8^@`

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`tan theta` `= 16/23`
`theta` `= 34.8^@`

  
`:.\ text(Angle of depression from)\ P\ text(to)\ Q = 34.8^@`

`text{(alternate,}\ PS\ text(||)\ RQ text{)}`

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`

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

♦ Mean mark 36%.

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

 
`=> D`

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 m
  2. 74 m
  3. 90 m
  4. 100 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`

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`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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a.    `53^circ`

b.    `190\ text(m)`

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♦♦ Mean mark (a) 31%
a.     `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^circ\ text{(nearest degree)}`

 

b.    `text(Find)\ CB:`

♦♦ Mean mark (b) 31%
MARKER’S COMMENT: The most efficient solution uses right-angled trigonometry.
`/_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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a.    `text{Proof  (See Worked Solutions)}`

b.    `17^@28^{′}`

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a.    `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.)`

 

b.    

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

♦♦ Mean mark 33%
MARKER’S COMMENT: If >30 “seconds”, round to the next “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`