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Measurement, STD2 M6 2019 HSC 22

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°4^{′}\ \ text{(nearest minute)}`

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`text(Using Pythagoras in)\ DeltaACD:`

Mean mark 51%.

`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, STD2 M6 2017 HSC 8 MC

The diagram shows a right-angled triangle.
 


 

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

  1. `70°16^{′}`
  2. ` 70°17^{′}`
  3. `70°27^{′}`
  4. `70°28^{′}`
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`B`

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`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 25b

2UG-2005-25b

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

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

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  1. `text(Proof)`
  2. `67°23^{′}`
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i.   `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”.

 
ii.  `sin ∠ABC = 12/13`

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