smc-4553-30-Sine Rule (Area)

Measurement, STD2 M6 2023 HSC 35

The diagram shows triangle `ABC`.

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

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`147\ text{m}^2`

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`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 rule (Area):}`

`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 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)

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Hence, find the area of the shaded region in square centimetres, correct to one decimal place. (3 marks)

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`13.86 \ text{cm}`
`45.1 \ text{cm}^2`

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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 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)

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

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`text(Find)\ AC:`

♦ Mean mark 39%.

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

`text(Using the sine 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 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`

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`D`

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`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 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)`

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`C`

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`text(Using sine 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 2010 HSC 26d

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

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`717\ text(m²)` `text{(nearest m²)}`

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♦♦ 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 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)`

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`C`

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`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`