Right-angled Triangles (Y12)

Measurement, STD1 M3 2025 HSC 22

An isosceles triangle is drawn inside a circle as shown. The base of the triangle is 4.8 cm long, the length of other sides is 4 cm and the height is $h$ cm.

Calculate the height, $h$, of triangle $ABC$. (2 marks)
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The area of triangle $ABC$ is 7.68 cm².
The radius of the circle is 2.5 cm.
Express the area of triangle $ABC$ as a percentage of the area of the circle, correct to 1 decimal place. (2 marks)
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a. $h=3.2\ \text{cm}$

b. $39.1\%$

Show Worked Solution

a. $\text{Since}\ \Delta ABC\ \text{is isosceles:}$

$BM\ \text{bisects}\ AC\ \ \Rightarrow\ \ AM=MC=2.4$

$\text{Using Pythagoras:}$

$h^2=4^2-2.4^2=16-5.76=10.24$

$h = \sqrt{10.24} = 3.2\ \text{cm}$

♦♦ Mean mark (a) 31%.

b. $\text{Area of circle}=\pi\times 2.5^2$

$\text{Percentage}$	$=\dfrac{\text{Area of triangle}}{\text{Area of circle}}\times 100\%$
	$=\dfrac{7.68}{\pi\times 2.5^2}\times 100\%$
	$=39.11\ …\%$
	$\approx 39.1\%\ \text{(1 decimal place)}$

♦♦♦ Mean mark (b) 22%.

Measurement, STD1 M3 2025 HSC 20

A map of a park containing a duck pond is shown.

A fence is built passing through the points $A$, $B$ and $C$ around the duck pond.

Using the scale provided on the map, calculate the length of the fence $AB$. (2 marks)
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The length of $AB$ is equal to the length of $BC$.
Use Pythagoras’ theorem to calculate the length of $AC$ in metres. Give your answer correct to 3 significant figures. (3 marks)
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What is the true bearing of point $A$ from point $C$ ? (2 marks)
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a. $75\ \text{metres}$

b. $106\ \text{metres}$

c. $225^\circ$

Show Worked Solution

a. $\text{Scale: 1 grid width = 5 metres}$

$AB = 15 \times 5 = 75\ \text{metres}$

Mean mark (a) 54%.

b. $\text{Using Pythagoras:}$

$AC^2=AB^2+BC^2$

$AC^2=75^2+75^2=11250$

	$\therefore\ AC$	$=\sqrt{11250}$
		$=106.066\ …$
		$\approx 106\ \text{m (3 sig fig)}$

♦♦ Mean mark (b) 38%.

c. $\text{Since }AB=BC:$

$\angle BAC=\angle BCA=45^\circ$

$\text{Bearing of \(A$ from $C$}\ =180+45=225^\circ\)

♦♦♦ Mean mark (c) 24%.

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$^\circ$, as shown.

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

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

Show Worked Solution

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

♦♦ Mean mark 34%.

Measurement, STD1 M3 2024 HSC 14

A hotel is located 186 m north and 50 m west of a train station.

What is the straight line distance from the hotel to the train station? Round your answer to the nearest metre. (2 marks)

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What is the bearing of the hotel from the train station? Round your answer to the nearest degree. (2 marks)

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a. $193\ \text{m}$

b. $345^\circ\ \text{(nearest degree)}$

Show Worked Solution

a. $\text{By Pythagoras:}$

	$d^2$	$=50^2+186^2$
	$d^2$	$=37\,096$
	$d$	$=\sqrt{37\,096}$
		$=192.603\ \dots$
		$=193\ \text{m (nearest metre)}$

♦♦ Mean mark (a) 37%.

b.	$\tan\theta$	$=\dfrac{50}{186}$
	$\theta$	$=15.046\ \dots^\circ$
		$\approx 15^\circ\ \text{(nearest degree)}$

$\text{Bearing}\ H\ \text{from}\ T =360-15=345^\circ\ \text{(nearest degree)}$

♦♦♦ Mean mark (b) 12%.

Measurement, STD1 M3 2024 HSC 7 MC

Consider the diagram shown.

Which of the following is the correct expression for the length of $x$ ?

$20\, \cos 40^{\circ}$
$20\, \sin 40^{\circ}$
$\dfrac{20}{\cos 40^{\circ}}$
$\dfrac{20}{\sin 40^{\circ}}$

Show Answers Only

$A$

Show Worked Solution

$\cos 40^{\circ}$	$=\dfrac{x}{20}$
$x$	$=20\, \times \cos 40^{\circ}$

$\Rightarrow A$

♦ Mean mark 49%.

Measurement, STD1 M3 2023 HSC 29

The diagram shows the location of three places $X$, $Y$ and $C$.

$Y$ is on a bearing of 120° and 15 km from $X$.

$C$ is 40 km from $X$ and lies due west of $Y$.

$P$ lies on the line joining $C$ and $Y$ and is due south of $X$.

Find the distance from $X$ to $P$. (2 marks)

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What is the bearing of $C$ from $X$, to the nearest degree? (2 marks)

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a. $7.5\ \text{km}$

b. $259^{\circ}$

Show Worked Solution

a. $\text{In}\ \Delta XPY:$

$\angle PXY=180-120=60^{\circ}$

$\cos 60^{\circ}$	$=\dfrac{XP}{15}$
$XP$	$=15\times \cos 60^{\circ}$
	$=7.5\ \text{km}$

♦♦ Mean mark (a) 24%.

b. $\text{In}\ \Delta XPC:$

$\text{Let}\ \theta = \angle CXP$

$\cos \theta$	$=\dfrac{7.5}{40}$
$\theta$	$=\cos^{-1} \Big(\dfrac{7.5}{40}\Big)$
	$=79.193\ …$
	$=79^{\circ}\ \text{(nearest degree)}$

$\text{Bearing}\ \ C\ \text{from}\ X=180+79=259^{\circ}$

♦♦♦♦ Mean mark (b) 9%.

Measurement, STD1 M3 2023 HSC 16

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)

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$369 \text{ m (nearest metre)}$

Show Worked Solution

$\text{Let }x\text{ be the distance from the cliff to the boat.}$

$\text{The angle at the boat}=18^{\circ}\text{ as it is alternate to the angle of depression from the cliff to the boat.}$

$\tan\theta$	$=\dfrac{\text{opp}}{\text{adj}}$
$\tan 18^{\circ}$	$=\dfrac{120}{x}$
$x$	$=\dfrac{120}{\tan 18^{\circ}}$
	$=369.322\ \ldots\ \text{m}$
	$\approx 369 \text{ m (nearest metre)}$

♦♦ Mean mark 30%.

Measurement, STD1 M3 2022 HSC 32

The diagram shows two right-angled triangles, `A B C` and `A B D`,

where `A C=35 \ text{cm}`, `B D=93 \ text{cm}`, `/_ A C B=41^@` and ` /_ A D B=\theta`.

Calculate the size of angle `\theta`, to the nearest minute. (4 marks)

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`19^@6^{′}`

Show Worked Solution

`text{In}\ Delta ABC:`

`tan 41^@`	`=(AB)/35`
`AB`	`=35xxtan 41^@`
	`=30.425\ …`

`text{In}\ Delta ABD:`

`sin theta`	`=(AB)/(BD)`
	`=(30.425\ …)/93`
`:.theta`	`=sin^(-1)((30.425\ …)/93)`
	`=19.09…`
	`=19^@6^{′}\ \ text{(nearest minute)}`

♦♦♦ Mean mark 17%.

Measurement, STD1 M3 2022 HSC 27

Shan is interested in buying a block of bushland. The price per hectare is $500. The land he wishes to purchase is in the shape of a right-angled triangle as shown.

The length of side `A B` is 7800 metres and the length of side `B C` is 3000 metres. The right angle of the triangle is angle `A C B`.

Note: `1\ \text(hectare) =10\ 000\ text(m)^2`

What is the cost of the block of bushland? (4 marks)

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`\$540\ 000`

Show Worked Solution

`text{Using Pythagoras:}`

`AC^2`	`=7800^2-3000^2`
	`=51\ 840\ 000`
`:.\ AC`	`=7200`

`text{Area}\ DeltaABC`	`=(BC xx AC)/2`
	`=(3000 xx 7200)/2`
	`=10\ 800\ 000\ text{m}^2`
	`=1080\ text{hectares (1 hectare = 10 000 m}^(2))`

`:.\ text{Cost}`	`=$500 xx 1080`
	`=$540\ 000`

♦♦ Mean mark 28%.

Measurement, STD1 M3 2022 HSC 8 MC

Which true bearing is the same as `text{S48°W}`?

`132^@`
`222^@`
`228^@`
`312^@`

Show Answers Only

`C`

Show Worked Solution

`text{True bearing}=180 + 48=228^@`

`=>C`

♦ Mean mark 47%.

Measurement, STD1 M3 2021 HSC 10 MC

The compass bearing of `B` from `A` is N38°E.

What is the true bearing of A from B?

128°
218°
232°
322°

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`B`

Show Worked Solution

♦♦♦ Mean mark 20%.

`text(Bearing)\ (A\ text(from)\ B)= 180 + 38= 218^@`

`=> B`

Measurement, STD1 M3 2021 HSC 28

A right-angled triangle `XYZ` is shown. The length of `XZ` is 16 cm and `angleYXZ = 30^@`.

Find the side length, `XY`, of the triangle in centimetres, correct to two decimal places. (2 marks)

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Hence, find the area of triangle `XYZ` in square centimetres, correct to one decimal place. (3 marks)

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a. `13.86\ text(cm)`

b. `55.4\ text(cm)^2`

Show Worked Solution

♦♦ Mean mark part (a) 24%.

a.	`cos 30^@`	`=(XY)/16`
	`XY`	`=16 xx cos30^@`
		`=13.856\ …=13.86\ text(cm)\ \ text{(2 d.p.)}`

b. `text(Using the sine rule:)`

♦♦♦ Mean mark part (b) 14%.

`text(Find )YZ`

`sin30^@`	`=(YZ)/16`
`YZ`	`=16\times sin30^@=8`

`text(Area)\ DeltaXYZ`	`= 1/2 xx8 xx 13.86`
	`=55.44=55.4\ text(cm)^2\ \ text{(1 d.p.)}`

Measurement, STD1 M3 2020 HSC 11

Consider the triangle shown.

Find the value of `theta`, correct to the nearest degree. (2 marks)

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Find the value of `x`, correct to one decimal place. (2 marks)

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a. `39^@`

b. `12.1 \ text{(to 1 d.p.)}`

Show Worked Solution

a.	`tan theta`	`= frac{8}{10}`
	`theta`	`= tan ^(-1) (frac{8}{10})`
		`= 38.659\ …`
		`= 39^@ \ text{(nearest degree)}`

♦ Mean mark 44% and 39% for part (a) and (b) respectively.

b. `text{Using Pythagoras:}`

`x`	`= sqrt{8^2 + 10^2}`
	`= 12.806\ …= 12.8 \ \ text{(to 1 d.p.)}`

Measurement, STD1 M3 2019 HSC 31

Two right-angled triangles, `ABC` and `ADC`, are shown.

Calculate the size of angle `theta`, correct to the nearest minute. (3 marks)

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`41^@5^{′}\ text{(nearest minute)}`

Show Worked Solution

`text(Using Pythagoras in)\ DeltaACD:`

♦♦♦ Mean mark 15%.

`AC^2`	`= 2.5^2 + 6^2`
	`= 42.25`
`:.AC`	`= 6.5\ text(cm)`

`text(In)\ DeltaABC:`

`costheta`	`= 4.9/6.5`
`theta`	`= cos^(−1)\ 4.9/6.5`
	`= 41.075\ …`
	`= 41^@4^{′}31^{″}`
	`= 41^@5^{′}\ text{(nearest minute)}`

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)

Show Answers Only

`32\ text(metres (nearest m))`

Show Worked Solution

`tan12^@ = h/150`

♦♦ Mean mark 33%.

`h`	`= 150 xx tan12^@`
	`= 31.88\ …`
	`= 32\ text(metres (nearest m))`

Measurement, STD1 M1 2019 HSC 4 MC

Which compass bearing is the same as a true bearing of 110°?

S20°E
S20°W
S70°E
S70°W

Show Answers Only

`C`

Show Worked Solution

`110° =\ text(S70°E)`

`=> C`

Measurement, STD2 M6 EQ-Bank 4

The diagram shows three checkpoints A, B and C. Checkpoint C is due east of Checkpoint A. The bearing of Checkpoint B from Checkpoint A is N22°E and the bearing of Checkpoint C from Checkpoint B is S68°E. The distance between Checkpoint A and Checkpoint B is 42 kilometres.

Mark the given information on the diagram and explain why `angleABC` is 90°. (2 marks)

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Find the distance, to the nearest kilometre, between Checkpoint A and Checkpoint C. (2 marks)

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If a runner is travelling 12.6 km/h, how long does it take her to travel between Checkpoint A and Checkpoint B, in hours and minutes? (2 marks)

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a. `text(See Worked Solutions)`

b. `112\ text{km}`

c. `3text(h 20 mins)`

Show Worked Solution

`angle ABC= 22 + 68= 90^@`

b. `text(In)\ \ DeltaABC,`

`cosangleBAC`	`= (AB)/(AC)`
`cos68°`	`= 42/(AC)`
`AC`	`= 42/(cos68°)`
	`= 112.11\ …= 112\ text{km (nearest km)}`

c.	`text(Travel time)`	`= text(dist)/text(speed)`
		`= 42/12.6`
		`= 3.333\ …= 3text(h 20 mins)`

Measurement, STD2 M6 EQ-Bank 7 MC

Jeet walks 5 km from his home on a bearing of 153°. He then walks due north until he arrives a point which is due east of his home.

How far east, to the nearest 0.1 km, is Jeet from home?

2.3 km
2.5 km
4.9 km
9.8 km

Show Answers Only

`A`

Show Worked Solution

`text(Jeet finishes at)\ P`

`text(Find)\ \ OP:`

`cos 63°`	`= (OP)/5`
`:. OP`	`= 5 xx cos 63°= 2.26\ …`

`=> A`

Measurement, STD2 M6 EQ-Bank 4 MC

Ralph travels from `P` to `Q` on a bearing of 130°. He then turns and walks to `R` on a bearing of 075°.

What is the size of `anglePQR`?

95°
100°
115°
125°

Show Answers Only

`D`

Show Worked Solution

`theta`	`= 90-40\ \ \ text{(180° in Δ)`
	`= 50^@`

`:. anglePQR= 50 + 75= 125^@`

`=>D`

Measurement, STD2 M6 EQ-Bank 5 MC

Which of the following expresses S65°W as a true bearing?

065°
155°
245°
295°

Show Answers Only

`C`

Show Worked Solution

`text(True bearing)= 180 + 65= 245°`

`=> C`

Measurement, STD2 M6 EQ-Bank 1 MC

In which triangle is `sin theta = 4/7`?

A.		B.
C.		D.

Show Answers Only

`D`

Show Worked Solution

`=>D`

Measurement, STD2 M6 2018 HSC 7 MC

The diagram shows the positions of towns `A`, `B` and `C`.

Town `A` is due north of town `B` and `angleCAB = 34°`

What is the bearing of town `C` from town `A`?

034°
146°
214°
326°

Show Answers Only

`C`

Show Worked Solution

`text(Bearing of Town)\ C\ text(from Town)\ A:`

`text(Bearing)= 180 + 34= 214^@`

`=>C`

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 2017 HSC 8 MC

The diagram shows a right-angled triangle.

What is the value of `theta`, to the nearest minute?

`70°16^{′}`
` 70°17^{′}`
`70°27^{′}`
`70°28^{′}`

Show Answers Only

`B`

Show Worked Solution

`tan theta`	`= text(opp)/text(adj)`
	`= 5.3/1.9= 2.789\ …`

COMMENT: An angle that has over 30″ (seconds) is rounded up to the next minute (i.e. rounded up to 70°17′).

`:. theta`	`= 70.277\ …^@`
	`=70°16^{′}39.8^{″}~~ 70^@17^{′}`

`=>B`

Measurement, STD2 M6 2005 HSC 8 MC

If `tan theta = 85`, what is the value of `theta`, correct to 2 decimal places?

`1.37°`
`1.56°`
`89.33°`
`89.20°`

Show Answers Only

`C`

Show Worked Solution

`tan theta`	`=85`
`theta`	`=tan^(-1)85=89.33°`

`=>C`

Measurement, STD2 M6 2016 HSC 26d

The diagram shows a block of land `ABCD` that has been surveyed. All measurements are in metres.

Calculate the length of `AB`, correct to the nearest metre. (2 marks)

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`46\ text{m}`

Show Worked Solution

`text(Using Pythagoras,)`

`AB`	`= sqrt(32^2 + 33^2)`
	`= 45.967\ …`
	`= 46\ text{m (nearest m)}`

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 m
74 m
90 m
100 m

Show Answers Only

`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 2005 HSC 25b

Use Pythagoras’ theorem to show that `ΔABC` is a right-angled triangle. (1 mark)

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Calculate the size of `∠ABC` to the nearest minute. (2 marks)

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a. `text(Proof)`

b. `67°23^{′}`

Show Worked Solution

a. `ΔABC\ text(is right-angled if)\ \ a^2 + b^2 = c^2`

`a^2 + b^2`	`= 5^2 + 12^2`
	`= 169`
	`= 13^2`
	`= c^2\ …\ text(as required.)`

MARKER’S COMMENT: Know your calculator process for producing an angle in minutes/seconds. Note >30 “seconds” rounds up to the higher “minute”.

b. `sin ∠ABC = 12/13`

`:.∠ABC`	`= 67.38\ …°`
	`=67°22^{′}48^{″}`
	`= 67°23^{′}\ \ \ text{(nearest minute)}`

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 Answers Only

`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 M1 2004 HSC 23a

The diagram shows the shape of Carmel’s garden bed. All measurements are in
metres.

Show that the area of the garden bed is 57 square metres. (2 marks)

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Carmel decides to add a 5 cm layer of straw to the garden bed.

Calculate the volume of straw required. Give your answer in cubic metres. (2 marks)

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Each bag holds 0.25 cubic metres of straw.

How many bags does she need to buy? (2 marks)

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A straight fence is to be constructed joining point A to point B.

Find the length of this fence to the nearest metre. (2 marks)

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a. `text(Proof)\ text{(See Worked Solutions)}`

b. `text(2.85 m)^3`

c. `text(She needs to buy 12 bags)`

d. `8\ text{m (nearest metre)}`

Show Worked Solution

a. `text(Area of)\ Delta ABC= 1/2 xx b xx h= 1/2 xx 10 xx 5.1=25.5\ text(m)^2`

`text(Area of)\ Delta ACD= 1/2 xx 10 xx 6.3= 31.5\ text(m)^2`

`:.\ text(Total Area)= 25.5 + 31.5= 57\ text(m)^2\ \text( … as required)`

b. `V= Ah= 57 xx 0.05= 2.85\ text(m)^3`

c. `text(Bags to buy)= 2.85/0.25= 11.4`

`:.\ text(She needs to buy 12 bags.)`

d. `text(Using Pythagoras,)`

`AB^2= 6.0^2 + 5.1^2= 36 + 26.01= 62.01`

`AB= 7.874\ …=8\ text{m (nearest metre)}`

Measurement, STD2 M6 2004 HSC 5 MC

What is the correct expression for tan 20° in this triangle?

`a/b`
`a/c`
`c/b`
`c/a`

Show Answers Only

`D`

Show Worked Solution

`tan 20^@= text(opposite)/text(adjacent)= c/a`

`=> D`

Measurement, STD2 M6 2007 HSC 8 MC

What is the length of the side `MN` in the following triangle, correct to two decimal places?

9.19 cm
10.07 cm
15.66 cm
18.67 cm

Show Answers Only

`C`

Show Worked Solution

`sin 50^@`	`= 12/(MN)`
`MN`	`= 12/sin50^@= 15.66\ …`

`=> C`

Measurement, STD2 M6 2008 HSC 14 MC

Danni is flying a kite that is attached to a string of length 80 metres. The string makes an angle of 55° with the horizontal.

How high, to the nearest metre, is the kite above Danni’s hand?

46 m
66 m
98 m
114 m

Show Answers Only

`B`

Show Worked Solution

`sin 55^@`	`= h/80`
`h`	`= 80 xx sin 55^@= 65.532\ …\ text(m)`

`=> B`

Measurement, STD2 M6 2014 HSC 26b

Calculate the value of `h` correct to two decimal places. (2 marks)

Show Answers Only

`10.65\ text{(2 d.p.)}`

Show Worked Solution

`sin 28^@`	`= 5/h`
`:. h`	`= 5/(sin 28^@)`
	`= 10.6502\ …`
	`= 10.65\ \ text{(2 d.p.)}`

Measurement, STD2 M6 2014 HSC 23 MC

The following information is given about the locations of three towns `X`, `Y` and `Z`:

• `X` is due east of `Z`

• `X` is on a bearing of 145° from `Y`

• `Y` is on a bearing of 060° from `Z`.

Which diagram best represents this information?

Show Answers Only

`C`

Show Worked Solution

♦ Mean mark 38%
COMMENT: Drawing a parallel North/South line through `Y` makes this question much simpler to solve.

`text(S)text(ince)\ X\ text(is due east of)\ Z`

`=> text(Cannot be)\ B\ text(or)\ D`

`text(The diagram shows we can find)`

`/_ZYX = 60 + 35^@ = 95^@`

`text(Using alternate angles)\ (60^@)\ text(and)`

`text(the)\ 145^@\ text(bearing of)\ X\ text(from)\ Y`

`=> C`

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) ---

Show Answers Only

a. `53^circ`

b. `190\ text(m)`

Show Worked Solution

♦♦ 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°.

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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a. `text{Proof (See Worked Solutions)}`

b. `17^@28^{′}`

Show Worked Solution

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.)`

`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 2009 HSC 4 MC

Which is the correct expression for the value of `x` in this triangle?

`8/cos30°`
`8/sin30°`
`8 xx cos30°`
`8 xx sin30°`

Show Answers Only

`A`

Show Worked Solution

`cos30^@`	`= 8/x`
`:.x`	`= 8/cos30^@`

`=> A`

Measurement, STD2 M6 2010 HSC 10 MC

A plane flies on a bearing of 150° from `A` to `B`.

What is the bearing of `A` from `B`?

`30^@`
`150^@`
`210^@`
`330^@`

Show Answers Only

`D`

Show Worked Solution

♦♦ Mean mark 34%

`/_TBA=30^@\ \ \ text{(angle sum of triangle)}`

`:.\ text(Bearing of)\ A\ text{from}\ B`

`=360-30=330^@`

`=> D`

Measurement, STD2 M6 2012 HSC 27d

A disability ramp is to be constructed to replace steps, as shown in the diagram.

The angle of inclination for the ramp is to be 5°.

Calculate the extra distance, `d`, that the ramp will extend beyond the bottom step.

Give your answer to the nearest centimetre. (3 marks)

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`386\ text(cm)`

Show Worked Solution

`text(Let the horizontal part of the ramp) = x\ text(cm)`

♦♦ Mean mark 35%
MARKER’S COMMENT: The better responses used a diagram of a simplified version of the ramp as per the Worked Solution.

`tan5^@`	`= 39/x`
`x`	`= 39/tan5^@`
	`= 445.772\ …`

`text(S)text(ince)\ \ x`	`= 60 + d`
`d`	`=445.772-60`
	`=385.772\ text(cm)`
	`=386\ text(cm)\ \ text{(nearest cm)}`

Measurement, STD2 M6 2011 HSC 9 MC

Two trees on level ground, 12 metres apart, are joined by a cable. It is attached 2 metres above the ground to one tree and 11 metres above the ground to the other.

What is the length of the cable between the two trees, correct to the nearest metre?

`9\ text(m)`
`12\ text(m)`
`15\ text(m)`
`16\ text(m)`

Show Answers Only

`C`

Show Worked Solution

`text(Using Pythagoras)`

`c^2`	`=12^2+9^2=144+81=225`
`c`	`=\sqrt{225}=15,\ \ c>0`

`=>C`

Measurement, STD2 M6 2010 HSC 3 MC

A field diagram has been drawn from an offset survey.

What is the distance from `G` to `H` correct to the nearest metre?

Show Answers Only

`B`

Show Worked Solution

`text(Using Pythagoras:)`

`GH^2=12^2+(16-11)^2=144+25=169`

`GH=sqrt169=13\ text(m)`

` => B`

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

Show Answers Only

`C`

Show Worked Solution

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

`=>C`

Measurement, STD2 M6 2012 HSC 4 MC

Which expression could be used to calculate the value of `x` in this triangle?

`29 xx cos40^@`
`29 xx cos 50^@`
`cos40^@/29`
`cos50^@/29`

Show Answers Only

`A`

Show Worked Solution

♦ Mean mark 42%

`cos40^@`	`= x/29`
`x`	`= 29xxcos40^@`

`=> A`

Measurement, STD2 M6 2013 HSC 4 MC

What is the value of `theta`, to the nearest degree?

`31^@`
`37^@`
`49^@`
`53^@`

Show Answers Only

`B`

Show Worked Solution

`sin theta`	`=81/135`
`:. theta`	`=37^@`

`=>\ B`

`text(Note that)\ tan theta\ text(or)\ cos theta\ text(could also be used.)`

\(\text{Percentage}\)	\(=\dfrac{\text{Area of triangle}}{\text{Area of circle}}\times 100\%\)
	\(=\dfrac{7.68}{\pi\times 2.5^2}\times 100\%\)
	\(=39.11\ …\%\)
	\(\approx 39.1\%\ \text{(1 decimal place)}\)

	\(\therefore\ AC\)	\(=\sqrt{11250}\)
		\(=106.066\ …\)
		\(\approx 106\ \text{m (3 sig fig)}\)

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

	\(d^2\)	\(=50^2+186^2\)
	\(d^2\)	\(=37\,096\)
	\(d\)	\(=\sqrt{37\,096}\)
		\(=192.603\ \dots\)
		\(=193\ \text{m (nearest metre)}\)

b.	\(\tan\theta\)	\(=\dfrac{50}{186}\)
	\(\theta\)	\(=15.046\ \dots^\circ\)
		\(\approx 15^\circ\ \text{(nearest degree)}\)

\(\cos 40^{\circ}\)	\(=\dfrac{x}{20}\)
\(x\)	\(=20\, \times \cos 40^{\circ}\)

\(\cos 60^{\circ}\)	\(=\dfrac{XP}{15}\)
\(XP\)	\(=15\times \cos 60^{\circ}\)
	\(=7.5\ \text{km}\)

\(\cos \theta\)	\(=\dfrac{7.5}{40}\)
\(\theta\)	\(=\cos^{-1} \Big(\dfrac{7.5}{40}\Big)\)
	\(=79.193\ …\)
	\(=79^{\circ}\ \text{(nearest degree)}\)

\(\tan\theta\)	\(=\dfrac{\text{opp}}{\text{adj}}\)
\(\tan 18^{\circ}\)	\(=\dfrac{120}{x}\)
\(x\)	\(=\dfrac{120}{\tan 18^{\circ}}\)
	\(=369.322\ \ldots\ \text{m}\)
	\(\approx 369 \text{ m (nearest metre)}\)