smc-1103-40-Angle of Elevation

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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\(\sin\ 4^\circ\)	\(=\dfrac{35}{p}\)
\(p\)	\(=\dfrac{35}{\sin\ 4^\circ}\)
	\(=501.74\dots\)
	\(= 502\ \text{m (nearest metre)}\)

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°.

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

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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{)}`