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Measurement, STD1 M3 2024 HSC 25

In a national park, a straight path connects a lookout to a car park. The lookout is 35 m higher than the car park. The path is inclined at an angle of elevation of 4°, as shown.
 

       

What is the length of the path, correct to the nearest metre?   (2 marks)

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\(502\ \text{m (nearest metre)}\)

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\(\sin\ 4^\circ\) \(=\dfrac{35}{p}\)
\(p\) \(=\dfrac{35}{\sin\ 4^\circ}\)
  \(=501.74\dots\)
  \(= 502\ \text{m (nearest metre)}\)
♦♦ Mean mark 34%.

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)

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`32\ text(metres  (nearest m))`

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`tan12° = h/150`

♦♦ Mean mark 33%.

`h` `= 150 xx tan12°`
  `= 31.88…`
  `= 32\ text(metres  (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^{′}`

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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{)}`