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) --- 5 WORK AREA LINES (style=lined) --- \(\text{Let }x\text{ be the distance from the cliff to the boat.}\) \(\text{The angle at the boat}=18^{\circ}\text{ as it is alternate to the angle of depression from the cliff to the boat.}\)
\(\tan\theta\)
\(=\dfrac{\text{opp}}{\text{adj}}\)
\(\tan 18^{\circ}\)
\(=\dfrac{120}{x}\)
\(x\)
\(=\dfrac{120}{\tan 18^{\circ}}\)
\(=369.322\ldots \text{m}\)
\(\approx 369 \text{ m (nearest metre)}\)
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?
- `60\ text(m)`
- `74\ text(m)`
- `90\ text(m)`
- `100\ text(m)`
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?
- 12.7 m
- 14.0 m
- 33.1 m
- 64.3 m
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.
- What is the angle of depression from `D` to `B`, correct to the nearest degree? (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
- How far did the boat travel from `C` to `B`, correct to the nearest metre? (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
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°.
- Show that the height of the building is approximately 19.5 m. (1 mark)
--- 2 WORK AREA LINES (style=lined) ---
- 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)
--- 5 WORK AREA LINES (style=lined) ---
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”.
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?
- 10 m
- 11 m
- 13 m
- 18 m