Non-right-angled Trig

Measurement, STD2 M6 2025 HSC 37

The diagram shows a park consisting of two equilateral triangles. The shaded triangle is a grassed section. All measurements on the diagram are in metres.

How long will it take to mow the grassed section if it takes 5 minutes to mow 20 m² ? Give your answer to the nearest minute. (4 marks)

--- 12 WORK AREA LINES (style=lined) ---

Show Answers Only

\(7 \ \text{minutes}\)

Show Worked Solution

♦♦ Mean mark 35%.

\(\text{Large triangle is equilateral (all sides = 12 m)}\)

\(\text{Area of large} \ \triangle\)	\(=\dfrac{1}{2}ab \, \sin C\)
	\(=\dfrac{1}{2} \times 12 \times 12 \times \sin 60^{\circ}\)
	\(=36 \sqrt{3}\)

\(\text{Area of} \ \ \triangle_1=\dfrac{1}{2} \times 3 \times 9 \times \sin 60^{\circ}=\dfrac{27 \sqrt{3}}{4}\)

\(\text{Grassed area}=36 \sqrt{3}-3 \times \dfrac{27 \sqrt{3}}{4}=27.2798 \ldots \ \text{m}^2\)

\(\text{Time to mow}\)	\(=\dfrac{27.2798 \ldots}{20} \times 5\)
	\(=6.81 \ldots\)
	\(=7 \ \text{minutes (nearest min)}\)

Measurement, STD2 M6 2025 HSC 35

The triangle \(PTA\) is shown. The length of \(PA\) is 75 m and the length of \(PT\) is 51 m.

The angle of depression from \(T\) to \(A\) is 36°, and the angle \(PTA\) is obtuse.

Find the length of \(TA\). Give your answer correct to 2 decimal places. (3 marks)

--- 10 WORK AREA LINES (style=lined) ---

Show Answers Only

\(TA=35.03 \ \text{m}\)

Show Worked Solution

♦♦♦ Mean mark 29%.

\(\angle TAP=36^{\circ} \ \text {(alternate)}\)

\(\text{Using sine rule in} \ \triangle TAP:\)

\(\dfrac{\sin \angle PTA}{75}\)	\(=\dfrac{\sin 36^{\circ}}{51}\)
\(\sin \angle PTA\)	\(=75 \times \dfrac{\sin 36^{\circ}}{54}=0.864 \ldots\)
\(\angle PTA\)	\(=\sin ^{-1}(0.864 \ldots)=180-59.81=120.19^{\circ}\ \ \text{(obtuse)}\)

\(\angle PTX=120.19-54=66.19^{\circ}\)

\(\angle TPA=90-66.19=23.81^{\circ}\ \left(180^{\circ}\ \text{in}\ \triangle \right)\)

\(\text{Using sine rule in} \ \triangle TAP:\)

\(\dfrac{TA}{\sin 23.81^{\circ}}\)	\(=\dfrac{51}{\sin 36^{\circ}}\)
\(TA\)	\(=\dfrac{51 \times \sin 23.81^{\circ}}{\sin 36^{\circ}}\)
	\(=35.03 \ \text{m (2 d.p.)}\)

Measurement, STD2 M6 2024 HSC 36

The diagram shows two vertical flagpoles, \(BE\) and \(CD\), set on sloping ground.

What is the height of the flagpole \(BE\), correct to 1 decimal place? (2 marks)
--- 6 WORK AREA LINES (style=lined) ---
What is the height of the flagpole \(CD\), correct to 1 decimal place? (2 marks)
--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

a. \(BE=25.4\ \text{m}\)

b. \(CD=19.7\ \text{m}\)

Show Worked Solution

a. \(\text{Using the sine rule:}\)

\(\dfrac{BE}{\sin 27^{\circ}}\)	\(=\dfrac{53.8}{\sin 106^{\circ}}\)
\(\therefore BE\)	\(=\dfrac{53.8 \times \sin 27^{\circ}}{\sin 106^{\circ}}\)
	\(=25.408…\)
	\(=25.4\ \text{m (1 d.p.)}\)

b. \(CD=EB-XB\)

\(\text{Consider}\ \Delta XBC:\)

\(\angle XBC=180-106=74^{\circ}, \ XC=ED=20\)

\(\tan 74^{\circ}\)	\(=\dfrac{20}{XB}\)
\(XB\)	\(=\dfrac{20}{\tan 74^{\circ}}\)
	\(=5.73\ \text{m}\)

\(CD=25.4-5.73=19.7\ \text{m (1 d.p.)}\)

♦♦ Mean mark (b) 35%.

Measurement, STD2 M6 2024 HSC 32

A regular pentagon \(ABCDE\) is drawn inside a circle with a radius 30 cm.

\(O\) is the centre of the circle.

What is the area of the shaded region of the circle. Answer correct to 2 significant figures. (4 marks)

--- 10 WORK AREA LINES (style=lined) ---

Show Answers Only

\(\text{690 cm}^{2}\)

Show Worked Solution

\(\text{Method 1}\)

\(\text{Area}\ \Delta ODC\)	\(= \dfrac{1}{2} ab \sin C\)
	\(=\dfrac{1}{2} \times 30 \times 30 \times \sin 72^{\circ} \)
	\(=427.98\ \text{cm}^{2}\)

\(\text{Shaded Area}\)	\(=\ \text{Area of circle}-5 \times\ \text{Area}\ \Delta ODC\)
	\(=(\pi \times 30^2)-(5 \times 427.98)\)
	\(=687.5\)
	\(=690\ \text{cm}^{2}\ \text{(2 sig.fig)}\)

♦ Mean mark 48%.

\(\text{Method 2}\)

\(\text{Consider}\ \Delta ODC:\)

\(\text{Area}=427.98\ \text{cm}^{2}\ \ \text{(see above)}\)

\(\text{Area of sector}\ ODC = \dfrac{72}{360} \times \pi \times 30^{2} = 565.49\ \text{cm}^{2}\)

\(\text{Shaded area (total)}\)	\(=(565.49-427.98) \times 5\)
	\(=690\ \text{cm}^{2}\ \text{(2 sig.fig)}\)

Measurement, STD2 M6 2023 HSC 35

The diagram shows triangle `ABC`.

Calculate the area of the triangle, to the nearest square metre. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`147\ text{m}^2`

Show Worked Solution

`text{Using the sine rule:}`

`(CB)/sin60^@`	`=12/sin25^@`
`CB`	`=sin60^@ xx 12/sin25^@`
	`=24.590…`

`angleACB=180-(60+25)=95^@\ \ text{(180° in Δ)}`

`text{Using the sine area rule:}`

`A`	`=1/2 xx AC xx CB xx sin angleACB`
	`=1/2 xx 12 xx 24.59 xx sin95^@`
	`=146.98…`
	`=147\ text{m}^2`

Measurement, STD2 M6 2023 HSC 33

The diagram shows a shape `APQBCD`. The shape consists of a rectangle `ABCD` with an arc `PQ` on side `AB` and with side lengths `BC` = 3.6 m and `CD` = 8.0 m.

The arc `PQ` is an arc of a circle with centre `O` and radius 2.1 m and `∠POQ=110°`.

What is the perimeter of the shape `APQBCD`? Give your answer correct to one decimal place. (4 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`23.8\ text{m}`

Show Worked Solution

`text{Arc length}\ PQ`	`=110/360 xx2 pi xx 2.1`
	`=4.03171… \ text{m}`

`text{Consider}\ ΔOPQ:`

♦ Mean mark 42%.

`sin 55^@`	`=x/2.1`
`x`	`=2.1 xx sin 55^@`
	`=1.7202…`

`PQ=2x=3.440\ text{m}`

`:.\ text{Perimeter}`	`=8+(2xx3.6)+4.031+(8-3.440…)`
	`=23.79…`
	`=23.8\ text{m (to 1 d.p.)}`

Measurement, STD2 M6 2022 HSC 26

The diagram shows two right-angled triangles, `ABC` and `ABD`,

where `AC=35 \ text{cm},BD=93 \ text{cm}, /_ACB=41^(@)` and `/_ADB=theta`.

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

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`19^@6^{′}`

Show Worked Solution

`text{In}\ Delta ABC:`

`cos 41^@`	`=35/(BC)`
`BC`	`=35/(cos 41^@)`
	`=46.375…`

`angle BCD = 180-41=139^@`

`text{Using sine rule in}\ Delta BCD:`

`sin theta/(46.375)`	`=sin139^@/93`
`sin theta`	`=(sin 139^@ xx 46.375)/93`
`:.theta`	`=sin^(-1)((sin 139^@ xx 46.375)/93)`
	`=19.09…`
	`=19^@6^{′}\ \ text{(nearest minute)}`

♦ Mean mark 50%.

Measurement, STD2 M6 2021 HSC 32

A right-angled triangle `XYZ` is cut out from a semicircle with centre `O`. The length of the diameter `XZ` is 16 cm and `angle YXZ` = 30°, as shown on the diagram.

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

--- 4 WORK AREA LINES (style=lined) ---
Hence, find the area of the shaded region in square centimetres, correct to one decimal place. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `13.86 \ text{cm}`

b. `45.1 \ text{cm}^2`

Show Worked Solution

a.	`cos 30^@`	`=(XY)/16`
	`XY`	`= 16 \ cos 30^@`
		`= 13.8564`
		`= 13.86 \ text{cm (2 d.p.)}`

b.	`text{Area of semi-circle}`	`= 1/2 times pi r^2`
		`= 1/2 pi times 8^2`
		`= 100.531 \ text{cm}^2`

♦ Mean mark part (b) 36%.

`text{Area of} \ Δ XYZ`	`= 1/2 ab\ sin C`
	`= 1/2 xx 16 xx 13.856 xx sin 30^@`
	`= 55.42 \ text{cm}^2`

`:. \ text{Shaded Area}`	`= 100.531-55.42`
	`= 45.111`
	`= 45.1 \ text{cm}^2 \ \ text{(1 d.p.)}`

Measurement, STD2 M6 2021 HSC 37

The diagram shows a triangle `ABC` where `AC` = 25 cm, `BC` = 16 cm, `angle BAC` = 28° and angle `ABC` is obtuse.

Find the size of the obtuse angle `ABC` correct to the nearest degree. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`133°`

Show Worked Solution

`text(Using the sine rule:)`

♦♦ Mean mark 31%.

`sin theta/25`	`= (sin 28°)/16`
`sin theta`	`= (25 xx sin 28°)/16`
`sin theta`	`= 0.73355`
`theta`	`= 47°`

`:. Obtuse angleABC`	`= 180-47`
	`= 133°`

Measurement, STD2 M6 EQ-Bank 24

The diagram shows a triangle with side lengths 25 cm and 47 cm and angle 30° and `theta`.

Find `theta` given it is an obtuse angle. Give your answer to the nearest minute. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`109°57^{′}`

Show Worked Solution

`text(Using sine rule:)`

`(sintheta)/47`	`= (sin30°)/25`
`sintheta`	`=47 xx (1/2)/25`
`sintheta`	`= 47/50`
`theta`	`= sin^(−1)\ 47/50`
	`= 70.05\ text(or)\ 109.94`
	`= 109.948…\ \ (theta\ text(is obtuse))`
	`= 109°57^{′}`

Measurement, STD2 M6 2019 HSC 17

The diagram shows a triangle with sides of length `x` cm, 11 cm and 13 cm and an angle of 80°.

Use the cosine rule to calculate the value of `x`, correct to two significant figures. (3 marks)

Show Answers Only

`16\ text(cm (2 sig. fig.))`

Show Worked Solution

`x^2`	`= 11^2 + 13^2-2 xx 11 xx 13 xx cos80°`
	`= 240.336…`
`:.x`	`= 15.502…`
	`= 16\ text(cm (2 sig. fig.))`

Measurement, STD2 M6 2018 HSC 30c

The diagram shows two triangles.

Triangle `ABC` is right-angled, with `AB = 13 text(cm)` and `/_ABC = 62°`.

In triangle `ACD, \ AD = x\ text(cm)` and `/_DAC = 40°`. The area of triangle `ACD` is 30 cm².

What is the value of `x`, correct to one decimal place? (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`8.1\ text{cm (1 d.p.)}`

Show Worked Solution

`text(Find)\ AC:`

♦ Mean mark 39%.

`sin62°`	`= (AC)/13`
`AC`	`= 13 xx sin62°`
	`= 11.478…`

`text(Using the sine area rule in)\ DeltaACD :`

`text(Area)`	`= 1/2 xx AC xx AD xx sin40°`
`30`	`= 1/2 xx 11.478… xx x xx sin40°`
`:.x`	`= (30 xx 2)/(11.478… xx sin40°)`
	`= 8.13…`
	`= 8.1\ text{cm (1 d.p.)}`

Measurement, STD2 M6 2018 HSC 12 MC

The diagram shows a triangle with side lengths 8 m, 9 m and 10m.

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

49°
51°
59°
72°

Show Answers Only

`text(D)`

Show Worked Solution

`text(Using the cosine rule:)`

`costheta`	`= (8^2 + 9^2-10^2)/(2 xx 8 xx 9)`
	`= 0.3125`
`:.theta`	`= cos^(−1)(0.3125)`
	`= 71.790…^@`

`=>D`

Measurement, STD2 M6 2015 HSC 30e

From point `S`, which is 1.8 m above the ground, a pulley at `P` is used to lift a flat object `F`. The lengths `SP` and `PF` are 5.4 m and 2.1 m respectively. The angle `PSC` is 108°.

Show that the length `PC` is 6.197 m, correct to 3 decimal places. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Calculate `h`, the height of the object above the ground. (4 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `6.197\ text{m (to 3 d.p.) … as required}`

b. `1.37\ text{m (to 2 d.p.)}`

Show Worked Solution

`text(Show)\ PC = 6.197\ text(m)`

♦ Mean mark (a) 41%.

`text(Using the cosine rule in)\ Delta PSC`

`PC^2`	`= PS^2 + SC^2-2 xx PS xx SC xx cos\ 108^@`
	`= 5.4^2 + 1.8^2-2 xx 5.4 xx 1.8 xx cos\ 108^@`
	`= 38.4072…`
`:.PC`	`= 6.19736…`
	`= 6.197\ text{m (to 3 d.p.) …as required}`

b. `text(Let)\ \ SD⊥PE`

♦♦ Mean mark (b) below 19%.
STRATEGY: Finding `PC` in part (i) and needing `PE` to find `h` should flag the strategy of finding `EC` and using Pythagoras.

`∠DSC\ text(is a right angle)`

`:.∠DSP = 108^@-90^@ = 18^@`

`text(In)\ ΔPDS`

`cos\ 18^@`	`= (DS)/5.4`
`DS`	`= 5.4 xx cos\ 18^@`
	`= 5.1357…\ text(m)`

`EC = DS = 5.1357…\ text{m (opposite sides of rectangle}\ DECS text{)}`

`text(Using Pythagoras in)\ Delta PEC:`

`PE^2 + EC^2 = PC^2`

`PE^2 + 5.1357^2`	`= 6.197^2`
`PE^2`	`= 12.027…`
`PE`	`= 3.468…\ text(m)`

`text(From the diagram,)`

`h`	`= PE-PF`
	`= 3.468…-2.1`
	`= 1.368…`
	`= 1.37\ text{m (to 2 d.p.)}`

Measurement, STD2 M6 2015 HSC 22 MC

The area of the triangle shown is 250 cm².

What is the value of `x`, correct to the nearest whole number?

`11`
`18`
`22`
`24`

Show Answers Only

`D`

Show Worked Solution

`text(Using)\ \ \ A = 1/2ab\ sin\ C`

♦ Mean mark 42%.

`250`	`= 1/2 xx 30x\ sin\ 44^@`
`250`	`= 15x\ sin\ 44 ^@`
`:.x`	`= 250/(15\ sin\ 44^@)`
	`= 23.99…\ text(m)`

`=>D`

Measurement, STD2 M6 2006 HSC 24b

A 130 cm long garden rake leans against a fence. The end of the rake is 44 cm from the base of the fence.

If the fence is vertical, find the value of `theta` to the nearest degree. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
The fence develops a lean and the rake is now at an angle of 53° to the ground. Calculate the new distance (`x` cm) from the base of the fence to the head of the rake. Give your answer to the nearest centimetre. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `text{70°}`

b. `text{109 cm}`

Show Worked Solution

`cos theta`	`= 44/130`
`theta`	`= 70.216… = 70^@\ text{(nearest degree)}`

`text(Using cosine rule:)`

`x^2`	`= 130^2 + 44^2-2 xx 130 xx 44 xx cos 53^@`
	`= 11\ 951.23…`
`x`	`= 109.32…= 109\ text{cm (nearest cm)}`

Measurement, STD2 M6 2006 HSC 9 MC

What is the area of this triangle, to the nearest square metre?

`text(152 m²)`
`text(283 m²)`
`text(328 m²)`
`text(351 m²)`

Show Answers Only

`C`

Show Worked Solution

`text(Using the Sine area rule)`

`A`	`= 1/2 ab\ sin C`
	`= 1/2 xx 39 xx 47 xx sin 21^@=\ text(328.44… m²)`

`=> C`

Measurement, STD2 M6 2005 HSC 5 MC

Which formula should be used to calculate the distance between Toby and Frankie?

`a/(sin A) = b/(sin B)`
`c^2 = a^2 + b^2`
`A = 1/2 ab\ sinC`
`c^2 = a^2 + b^2 − 2ab\ cosC`

Show Answers Only

`A`

Show Worked Solution

`text(The triangle is not a right-angled triangle,)`

`:.\ text(Not)\ B`

`text(Given the information on the diagram provides)`

`text(2 angles and 1 side, the sine rule will work best.)`

`a/sinA = b/sinB`

`=> A`

Measurement, STD2 M6 2004 HSC 9 MC

What is the area of the triangle to the nearest square metre?

`text(102 m²)`
`text(153 m²)`
`text(172 m²)`
`text(178 m²)`

Show Answers Only

`C`

Show Worked Solution

`text(Using sine area rule,)`

`text(Area)`	`= 1/2 ab sin C`
	`= 1/2 xx 30 xx 20 xx sin 35^@=172.072…\ text(m²)`

`=> C`

Measurement, STD2 M6 2007 HSC 25b

The angle of depression from `J` to `M` is 75°. The length of `JK` is 20 m and the length of `MK` is 18 m.

Copy or trace this diagram into your writing booklet and calculate the angle of elevation from `M` to `K`. Give your answer to the nearest degree. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`58^@`

Show Worked Solution

`/_AJL = 90^@`

`/_MJL = 90-75 = 15^@`

`text(Using sine rule in)\ Delta MJK`

`text(Let)\ /_JMK = x^@`

`20/sin x`	`= 18/sin15^@`
`18 sin x`	`= 20 xx sin 15^@`
`sin x`	`= (20 xx sin 15^@)/18 = 0.2875…`
`x`	`= 16.71…^@`

`/_JML = 75^@\ text{(} text(alternate angles,)\ ML \ text(||) \ AJ text{)}`

`:.\ /_KML`	`= 75^@-16.71…`
	`= 58.287…^@`
	`= 58^@\ \ \ text{(nearest degree)}`

`:.\ text(Angle of Elevation from)\ M\ text(to)\ K\ text(is)\ 58^@.`

Measurement, STD2 M6 2008 HSC 25c

Pieces of cheese are cut from cylindrical blocks with dimensions as shown.

Twelve pieces are packed in a rectangular box. There are three rows with four pieces of cheese in each row. The curved surface is face down with the pieces touching as shown.

What are the dimensions of the rectangular box? (4 marks)

--- 8 WORK AREA LINES (style=lined) ---

To save packing space, the curved section is removed.
What is the volume of the remaining triangular prism of cheese? Answer to the nearest cubic centimetre. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `41\ text(cm) xx 21\ text(cm) xx 15\ text(cm)`

b. `506\ text(cm)^3`

Show Worked Solution

a. `text(Box height) = 15\ text(cm)`

♦ Mean mark (a) 45%.

`text{(radius of the arc)}`

`text(Box width)`	`= 3 xx 7= 21\ text(cm)`
`text(Box length)`	`= 4x`

`text(Using cosine rule:)`

`c^2`	`= a^2 + b^2-2ab\ cos C`
`x^2`	`= 15^2 + 15^2-2 xx 15 xx 15 xx cos 40^@`
	`= 450-344.7199…= 105.2800…`
`x`	`= 10.2606…`

`text{Box length} = 4 xx 10.2606…= 41.04…`

`:.\ text(Dimensions are)\ \ 41\ text(cm) xx 21\ text(cm) xx 15\ text(cm)`

b. `text(Volume) = Ah`

♦♦♦ Mean mark (b) 22%.

`h = 7\ text(cm)`

`A= 1/2 ab\ sin C= 1/2 xx 15 xx 15 xx sin 40^@= 72.3136…`

`:. V`	`= 72.3136… xx 7= 506.195…`
	`= 506\ text(cm)^3\ \ text{(nearest whole)}`

Measurement, STD2 M6 2008 HSC 5 MC

What is the size of the smallest angle in this triangle?

`29^@`
`47^@`
`58^@`
`76^@`

Show Answers Only

`B`

Show Worked Solution

`text(Smallest angle is opposite smallest side.)`

`cos A`	`= (b^2 + c^2-a^2)/(2bc)`
	`= (7^2 + 8^2-6^2)/(2 xx 7 xx 8)`
	`= 0.6875`
`A`	`=cos ^(-1)(0.6875)`
`:.\ A`	`= 46.567…^@`

`=> B`

Measurement, STD2 M6 2010 HSC 26d

Find the area of triangle `ABC`, correct to the nearest square metre. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`717\ text(m²)` `text{(nearest m²)}`

Show Worked Solution

♦♦ Mean mark 32%.
TIP: The allocation of 3 marks to this question should flag the need for more than 1 step.

`cos/_C`	`=(AC^2 + CB^2-AB^2)/(2 xx AC xx CB)`
	`=(50^2 + 40^2-83^2)/(2 xx 50 xx 40)`
	`= -0.69725…`
`/_C`	`=134.2067…^@`

`text(Using Area) = 1/2 ab\ sinC :`

`text(Area)\ Delta ABC`	`=1/2 xx 50 xx 40 xx sin134.2067…^@`
	`=716.828…`
	`=717\ text(m²)\ \ \ \ text{(nearest m²)}`

Measurement, STD2 M6 2013 HSC 26a

Triangle `PQR` is shown.

Find the size of angle `Q`, to the nearest degree. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`110^@\ \ \ text{(nearest degree)}`

Show Worked Solution

♦ Mean mark 47%

`text(Using Cosine rule)`

`cos /_Q`	`= (a^2 + b^2-c^2)/(2ab)`
	`= (53^2 + 66^2-98^2)/(2xx53xx66)`
	`=-0.3486…`

`:. /_Q`	`= 110.4034…`
	`= 110^@\ \ \ text{(nearest degree)}`

Measurement, STD2 M6 2009 HSC 22 MC

In the diagram, `AD` and `DC` are equal to 30 cm.

What is the length of `AB` to the nearest centimetre?

`28\ text(cm)`
`31\ text(cm)`
`34\ text(cm)`
`39\ text(cm)`

Show Answers Only

`A`

Show Worked Solution

♦ Mean mark of 35%

`Delta ADC\ text(is isosceles)`

`/_DAB = /_DCA`	`= x^@`
`2x + 80^@`	`= 180^@\ \ \ (text{Angle sum of}\ DeltaADC)`
`2x`	`= 100^@`
`x`	`= 50^@`

`/_ DBA`	`= 180\-(50 + 60)\ \ \ (text{Angle sum of}\ Delta ADB)`
	`= 70^@`

`text(Using sine rule:)`

`(AB)/sin60`	`= 30/sin70`
`AB`	`= (30 xx sin60)/sin70`
	`= 27.648…\ text(cm)`

`=> A`

Measurement, STD2 M6 2010 HSC 9 MC

Three towns `P`, `Q` and `R` are marked on the diagram.

The distance from `R` to `P` is 76 km. `angle RQP=26^circ` and `angle RPQ=46^@.`

What is the distance from `P` to `Q` to the nearest kilometre?

`100\ text(km)`
`125\ text(km)`
`165\ text(km)`
`182\ text(km)`

Show Answers Only

`C`

Show Worked Solution

`angle QRP`	`=180-(26+46) (180^circ\ text(in) \ Delta)`
	`=108^circ`

`text{Using sine rule}`

`(PQ)/sin108^circ`	`=76/sin26^circ`
`PQ`	`=(76xxsin108^circ)/sin26^circ=164.88\ text(km)`

`=> C`

Measurement, STD2 M6 2012 HSC 29c

Raj cycles around a course. The course starts at `E`, passes through `F`, `G` and `H` and finishes at `E`. The distances `EH` and `GH` are equal.

What is the length of `EF`, to the nearest kilometre? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
What is the total distance that Raj cycles, to the nearest kilometre? (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `22\ text{km}`

b. `202\ text{km}`

Show Worked Solution

a. `text(Find)\ EF:`

♦ Mean mark 48%.

`/_ FGE= 180\-(139 + 31)= 10^@ \ \ text{(angle sum of}\ Delta EFGtext{)}`

`text(Using Sine rule:)`

`(EF)/sin10^@`	`= 82/sin139^@`
`EF`	`= (82 xx sin10^@)/sin139^@= 21.70406…= 22\ text{km (nearest km)}`

b. `text(Let)\ \ d = text(total distance cycled)`

`text(Find)\ EH:`

`text(S)text(ince)\ Delta EGH\ text(is isosceles, and)\ /_EHG = 90^@`

`/_GEH = /_HGE = 45^@`

`text{(angles opposite equal sides in}\ Delta EGHtext{)}`

♦ Mean mark 37%.
MARKER’S COMMENT: Students could also have used Pythagoras or the Sine rule to calculate `GH`.

`sin45^@`	`= (GH)/82`
`GH`	`= 82 xx sin45^@= 57.983…`

`:. d`	`= EF + FG + GH + EH`
	`= 21.704… + 64 + 57.983… + 57.983…`
	`= 201.66…`
	`= 202 \ text{km (nearest km)}`

Measurement, STD2 M6 2012 HSC 10 MC

What is the area of this triangle, to the nearest square metre?

`33\ text(m²)`
`37\ text(m²)`
`42\ text(m²)`
`44\ text(m²)`

Show Answers Only

`C`

Show Worked Solution

`text(Let unknown angle)=/_C`

`/_C`	`= 180-(50 + 57)\ \ \ \ \ (180^@ \ text(in)\ Delta)`
	`=73^@`

`:. A`	`= 1/2 ab\ sinC`
	`= 1/2 xx 9.9 xx 8.8 xx sin73^@= 41.656 \ text(m²)`

`=> C`

Measurement, STD2 M6 2013 HSC 24 MC

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

`21^@`
`32^@`
`43^@`
`55^@`

Show Answers Only

`C`

Show Worked Solution

`a/sinA`	`=b/sinB`
`82/sinA`	`=100/sin26`
`sin A`	`=(82 xx sin26)/100`
	`=0.35946…`
`/_A`	`=21^@\ \ \ \ text{(nearest degree)}`

`text(S)text(ince)\ 180^@\ text(in)\ Delta:`

`90+26+(theta+21)`	`=180`
`theta`	`=43^@`

`=> C`