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Trigonometry, SMB-039

Gerry flies his helicopter from a heliport on a bearing of S50°W for 75 km until he stops at Town P which is due west of Town Q.

If Town Q is due south of the heliport, find the distance between the heliport and Town Q in kilometres, correct to two decimal places.   (3 marks)

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\(48.21\ \text{km} \)

Show Worked Solution

\(\text{Let}\ \ x=\ \text{distance from heliport and Town Q}\)

\(\cos 50^{\circ}\) \(= \dfrac{x}{75} \)  
\(x\) \(= 75 \times \cos 50^{\circ}\)  
  \(= 48.209…\)  
  \(= 48.21\ \text{km} \)  

Trigonometry, SMB-038

Boat A leaves port on a bearing of 065° and sails for 22 km until it is due east of Boat B. 

If Boat B is due north of the port, find the distance between the port and Boat B in kilometres, correct to two decimal places.   (3 marks)

Show Answers Only

\(9.30\ \text{km} \)

Show Worked Solution
 

\(\text{Let}\ \ x=\ \text{distance from port to Boat B}\)

\(\cos 65^{\circ}\) \(= \dfrac{x}{22} \)  
\(x\) \(= 22 \times \cos 65^{\circ}\)  
  \(= 9.2976…\)  
  \(= 9.30\ \text{km} \)  

Trigonometry, SMB-035 MC

Peter is standing 8 metres due west of Mary.

Bob is standing 8 metres due north of Mary.

Bob is facing north. He then turns anticlockwise so that he is facing Peter.

How many degrees does Bob turn through?

  1. `45^@`
  2. `60^@`
  3. `120^@`
  4. `135^@`
Show Answers Only

 `D`

Show Worked Solution

`text(Degrees Bob turns through)`

`= 90 + 45`

`= 135^@`

`=>D`

Trigonometry, SMB-031

  1. Express the ratio of `tan theta`  as a fraction.  (2 marks)

  2. Hence or otherwise, find the value of `theta`, correct to two decimal places.  (1 mark)

Show Answers Only
  1. `tan theta=15/8`
  2. `61.93^@`
Show Worked Solution

i.    `text{By Pythagoras,}`

`x^2+8^2` `=17^2`  
`x^2` `=289-64`  
  `=225`  
`x` `=15`  

 
`:.tan theta = 15/8`

 

ii.   `theta` `=tan^{-1}(15/8)`  
  `=61.927…^@`  
  `=61.93^@\ text{(to 2 d.p.)}`  

Trigonometry, SMB-030

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

Show Answers Only

`54^@ 28^{′}`

Show Worked Solution
`tan theta` `=7/5`  
`theta` `=tan^{-1}(7/5)`  
  `=54.462…`  
  `=54^@ 27^{′}44^{″}`  
  `=54^@ 28^{′}`  
NOTE: Seconds > 30 are rounded up to the higher minute.

Trigonometry, SMB-029

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

Show Answers Only

`61^@ 56^{′}`

Show Worked Solution
`tan theta` `=15/8`  
`theta` `=tan^{-1}(15/8)`  
  `=61.927…`  
  `=61^@55^{′}39^{″}`  
  `=61^@ 56^{′}`  
NOTE: Seconds > 30 round up to the higher minute.

Trigonometry, SMB-024

Express `tan theta` as a fraction in its simplest form.  (3 marks)

Show Answers Only

`tan theta = (2sqrt5)/5`

Show Worked Solution

`text{Let unknown side =}\ x`

`text{By Pythagoras:}`

`6^2` `=x^2 + 4^2`  
`x^2` `=36-16`  
`x` `=sqrt20\ \ (x>0)`  

 

`:.tan theta` `=4/sqrt20`  
  `= 4/(2sqrt5) xx sqrt5/sqrt5`  
  `= (2sqrt5)/5`  

Trigonometry, SMB-023

Using Pythagoras and showing your working, express `tan 30^{@}` as a fraction.  (2 marks)

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`tan 30^{@} = 1/sqrt3`

Show Worked Solution

`text{Let}\ \ x =\ text{unknown side}`

`text{By Pythagoras:}`

`2^2` `=x^2 + 1^2`  
`x^2` `=4-1`  
`x` `=sqrt3\ \ (x>0)`  

 
`tan 30^{@} = text{opp}/text{adj} = 1/sqrt3`

Trigonometry, SMB-022

Find the value of `theta`, correct to the nearest minute.  (3 marks)

Show Answers Only

`66^@ 25^{′}`

Show Worked Solution

`cos theta` `=2/5`  
`theta` `=cos^{-1}(2/5)`  
  `=66.421…^@`  
  `=66^@ 25^{′} text{(to 1 d.p.)}`  

Trigonometry, SMB-021

  1. Express the ratio of `cos theta`  in its simplest form.  (2 marks)

  2. Hence, find the value of `theta`, correct to two decimal places.  (1 mark)

Show Answers Only
  1. `cos theta = 4/5`
  2. `theta = 36.87^@`
Show Worked Solution

i.    `text{By Pythagoras,}`

`x^2+8^2` `=10^2`  
`x^2` `=100-36`  
  `=64`  
`x` `=8`  

 
`:.cos theta = 8/10=4/5`

 

ii.   `theta` `=cos^{-1}(4/5)`  
  `=36.869…^@`  
  `=36.87^@\ text{(to 2 d.p.)}`  

Trigonometry, SMB-020


  

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

Show Answers Only

`70^@ 32^{′}`

Show Worked Solution
`cos theta` `=4/12`  
`theta` `=cos^{-1}(1/3)`  
  `=70.528…`  
  `=70^@ 31^{′}43^{″}`  
  `=70^@ 32^{′}`  
NOTE: Seconds > 30 are rounded up to the higher minute.

Trigonometry, SMB-014

Express `cos theta` as a fraction in its simplest form.  (3 marks)

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`cos theta = sqrt3/2`

Show Worked Solution

`text{Let unknown side =}\ x`

`text{By Pythagoras:}`

`4^2` `=x^2 + 2^2`  
`x^2` `=16-4`  
`x` `=sqrt12\ \ (x>0)`  

 

`:.cos theta` `=sqrt12/4`  
  `= (2sqrt3)/4`  
  `= sqrt3/2`  

Trigonometry, SMB-013

Using Pythagoras and showing your working, express `cos 30^{@}` as a fraction.  (2 marks)

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`cos 30^{@} = sqrt3/2`

Show Worked Solution

`text{Let}\ \ x =\ text{unknown side}`

`text{By Pythagoras:}`

`2^2` `=x^2 + 1^2`  
`x^2` `=4-1`  
`x` `=sqrt3\ \ (x>0)`  

 
`cos 30^{@} = text{adj}/text{hyp} = sqrt3/2`

Trigonometry, SMB-004

Without the use of a calculator, express the following as a fraction

  1. `sin 30^{@}`   (1 mark)

  2. `sin 60^{@}`   (2 marks)

Show Answers Only
  1. `sin 30^{@} = 1/2`
  2. `sin 60^{@} = sqrt3/2`
Show Worked Solution

i.    `sin 30^{@} = text{opp}/text{hyp} = 1/2`

ii.   `text{Let}\ \ x =\ text{unknown side}`

`text{By Pythagoras:}`

`2^2` `=x^2 + 1^2`  
`x^2` `=4-1`  
`x` `=sqrt3\ \ (x>0)`  

 
`sin 60^{@} = text{opp}/text{hyp} = sqrt3/2`

Trigonometry, SMB-003

Express the following as a fraction

  1. `sin theta`  (1 mark)

  2. `tan theta`  (1 mark)

Show Answers Only
  1. `sin theta = 7/25`
  2. `tan theta = 7/24`
Show Worked Solution

i.    `sin theta = text{opp}/text{hyp} = 7/25`

ii.   `tan theta = text{opp}/text{adj} = 7/24`

Trigonometry, SMB-002

Express the following as a simplified fraction

  1. `cos theta`  (1 mark)

  2. `tan theta`  (1 mark)

Show Answers Only
  1. `cos theta = 3/5`
  2. `tan theta = 4/3`
Show Worked Solution

i.    `cos theta = text{adj}/text{hyp} = 6/10 = 3/5`

ii.   `tan theta = text{opp}/text{adj} = 8/6 = 4/3`

Trigonometry, SMB-001

Express the following as a fraction

  1. `cos theta`  (1 mark)

  2. `sin theta`  (1 mark)

Show Answers Only
  1. `cos theta = 12/13`
  2. `sin theta = 5/13`
Show Worked Solution

i.    `cos theta = text{adj}/text{hyp} = 12/13`

ii.   `sin theta = text{opp}/text{hyp} = 5/13`

Measurement, STD2 M6 2020 HSC 16

Consider the triangle shown.
 


 

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

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

Show Answers Only
  1. `39^@`
  2. `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)}`

 

b.     `text{Using Pythagoras:}`

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

Measurement, STD2 M6 2019 HSC 12 MC

An owl is 7 metres above ground level, in a tree. The owl sees a mouse on the ground at an angle of depression of 32°.

How far must the owl fly in a straight line to catch the mouse, assuming the mouse does not move?

  1.  3.7 m
  2.  5.9 m
  3.  8.3 m
  4.  13.2 m
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ OM = text(Flight distance)`

♦ Mean mark 36%.

`sin32°` `= 7/(OM)`
`:. OM` `= 7/(sin32°)`
  `= 13.2\ text(m)`

 
`=> D`

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)

Show Answers Only

`41°4^{′}\ \ text{(nearest minute)}`

Show Worked Solution

`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 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 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?

  1. `60\ text(m)`
  2. `74\ text(m)`
  3. `90\ text(m)`
  4. `100\ text(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 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?
 


 

  1. 12.7 m
  2. 14.0 m
  3. 33.1 m
  4. 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 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.
 

  1. What is the angle of depression from `D` to `B`, correct to the nearest degree?   (3 marks)

  2. How far did the boat travel from `C` to `B`, correct to the nearest metre?   (2 marks)

Show Answers Only
  1. `53^circ`
  2. `190\ text(m)`
Show Worked Solution
♦♦ Mean mark 31%
i.    `tan/_ADB` `=126/168`
  ` /_ADB` `=36.8698…`
    `=36.9^circ\ \ \ \ text{(to 1 d.p)}` 

 

`/_text(Depression)\ D\ text(to)\ B` `=90-36.9`
  `=53.1`
  `=53^circ\ text{(nearest degree)}`

 

ii.     `text(Find)\ CB:`

♦♦ Mean mark 31%
MARKER’S COMMENT: Solve efficiently by using right-angled trigonometry. Many students used non-right angled trig, adding to the calculations and the difficulty.
`/_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°.
 

  1. Show that the height of the building is approximately 19.5 m.   (1 mark)

  2. 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)

Show Answers Only

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

ii.   `17°28^{′}`

Show Worked Solution

i.  `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.)`

 

ii.  

`text(Let)\ \ /_ \ text(Elevation (from car) ) = theta`

♦♦ Mean mark 33%
MARKER’S COMMENT: If >30 “seconds”, round to the higher “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? 
 

 

  1. `8/cos30°` 
  2. `8/sin30°` 
  3. `8 xx cos30°`  
  4. `8 xx sin30°` 
Show Answers Only

`A`

Show Worked Solution
`cos30^@`  `= 8/x`
`:.x` `= 8/cos30^@` 

 
`=>  A`

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)

Show Answers Only

 `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 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?

  1. 10 m
  2. 11 m
  3. 13 m
  4. 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? 
 

 

  1. `29 xx cos40^@`  
  2. `29 xx cos 50^@`  
  3. `cos40^@/29`  
  4. `cos50^@/29`  
Show Answers Only

`A`

Show Worked Solution
♦ Mean mark 42%
`cos40^@` `= x/29`
`x` `= 29xxcos40^@`

 
`=>  A`