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Trigonometry, 2ADV T1 2021 HSC 12

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.
 


 

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

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

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  1. `13.86 \ text{cm}`
  2. `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`

 

`text{Area of} \ Δ XYZ` `= 1/2 a b 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.)}`  

Trigonometry, 2ADV T1 2021 HSC 18

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)

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`133°`

Show Worked Solution

`text(Using the sine rule:)`

`sin theta/25` `= (sin 28°)/16`
`sin theta` `= (25 xx sin 28°)/16`
`sin theta` `= 0.73355`
`theta` `= 47°`
 
`:. angleABC= 180-47= 133°`

Trigonometry, 2ADV T1 EQ-Bank 2

Determine all possible dimensions for triangle `ABC` given  `AB = 6.2\ text(cm)`, `angleABC = 35°`  and  `AC = 4.1`.

Give all dimensions correct to one decimal place.  (3 marks)

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`text(7.1 cm, 6.2 cm, 4.1 cm or)`

`text(3.0 cm, 6.2 cm, 4.1 cm.)`

Show Worked Solution

`text(Using the sine rule:)`

`(sinangleACB)/6.2` `= (sin35^@)/4.1`
`sinangleACB` `= (6.2 xx sin35^@)/4.1`
  `= 0.8673…`
`angleACB` `= 60.15…^@\ text(or)\ 119.84…^@`

  
`text(If)\ \ angleACB = 60.15^@,`

`angleBAC = 180-(35 + 60.15) = 84.85^@`
 

`(BC)/(sin84.85)` `= 4.1/(sin35^@)`
`BC` `= 7.11… = 7.1\ text(cm)`

 
`text(If)\ \ angleACB = 119.85^@,`

`angleBAC = 180-(35 + 119.85) = 25.15^@`
 

`(BC)/(sin25.15)` `= 4.1/(sin35^@)`
`BC` `= 3.03… = 3.0\ text(cm)`

 
`:.\ text(Possible dimensions are:)`

`text(7.1 cm, 6.2 cm, 4.1 cm or)`

`text(3.0 cm, 6.2 cm, 4.1 cm.)`

Trigonometry, 2ADV T1 2006 HSC 1d

2006 1d

 
Find the value of `theta` in the diagram. Give your answer to the nearest degree.  (2 marks)

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`18°`

Show Worked Solution

`text(Using the sine rule)`

`sin theta / 5` `= (sin 33°)/9`
`sin theta` `= (5 xx sin 33°)/9`
  `= 0.30257…`
`:. theta` `= 17.612…`
  `= text{18°  (nearest degree)}`

Trigonometry, 2ADV T1 2013 HSC 14c

2013 14c

The right-angled triangle `ABC` has hypotenuse  `AB = 13`. The point `D` is on `AC` such that  `DC = 4`, `/_DBC = pi/6`  and  `/_ABD = x`.

Using the sine rule, or otherwise, find the exact value of  `sin x`.   (3 marks)

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 `(7sqrt3)/26 text(.)`

Show Worked Solution

`text(Find)\ \ /_ADB:`

`/_ADB` `= pi/6 + pi/2 \ \ \ text{(exterior angle of}\ Delta BDC text{)}`
  `= (2pi)/3\ text(radians)`

 
`text(Find)\ \ AD`

♦ Mean mark 36%.
STRATEGY TIP: The hint to use the sine rule should flag to students that they will be dealing in non-right angled trig (i.e. `Delta ABD`) and to direct their energies at initially finding `/_ADB` and `AD`.
`tan (pi/6)` `= 4/(BC)`
`1/sqrt3` `=4/(BC)`
`BC` `=4 sqrt3`

 

`text(Using Pythagoras:)`

`AC^2 + BC^2` `= AB^2`
`AC^2 + (4sqrt3)^2` `= 13^2`
`AC^2` `= 16-48= 121`
`=>AC` `= 11`

 
`:.AD=AC-DC= 11-4=7`
 

`text(Using sine rule:)`

`(AB)/(sin /_BDA)` `= (AD)/(sinx)`
`13/(sin ((2pi)/3))` `=7/(sinx)`
`13 xx sinx` `= 7 xx sin ((2pi)/3)`
`sinx` `= 7/13 xx sin((2pi)/3)`
  `= 7/13 xx sqrt3/2`
  `= (7 sqrt3)/26`

 
`:.\ text(The exact value of)\ sinx = (7sqrt3)/26 text(.)`