smc-6928-20-Right-Angled Trig

Measurement, STD2 2026 HSC 34

The diagram shows a right-angled triangle $OAB$ and a circle with centre $O$ and radius $r$ cm, where $OA$ and $OB$ cut the circle at $X$ and $Y$ respectively and $AB=5$ cm, $OB=12$ cm and $AX=7$ cm. The part of the triangle outside the circle is shaded.

What percentage of the triangle is shaded, correct to 1 decimal place? (5 marks)

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$76.3\%$

Show Worked Solution

$\text{Find $r$ (by Pythagoras):}$

$OA=\sqrt{12^2+5^2}=13$

$r=13-7=6\ \text{cm}$

$\text{Find}\ \angle AOB:$

$\tan \angle AOB=\dfrac{5}{12}$

$\angle AOB = \tan^{-1}\left(\dfrac{5}{12}\right)=22.619\ldots^{\circ}$

$\text{Sector area ($OXY$)}=\dfrac{\theta} {360^{\circ}}\times\pi r^2=\dfrac{22.6\ldots} {360^{\circ}}\times\pi(6)^2=7.10624\ldots\ \text{cm}^2$

$\text{Triangle area}=\dfrac{1}{2}\times12\times5=30\ \text{cm}^2$

$\text{Shaded triangle area}=30-7.1\ldots = 22.89\dots \text{cm}^2$

$\text{Percentage shaded}=\dfrac{22.89\dots}{30}\times100\% = 76.3\%\ \text{(1 d.p.)}$

Measurement, STD2 M6 2024 HSC 6 MC

Consider the diagram shown.

Which of the following is the correct expression for the length of $x$ ?

$20\, \cos 40^{\circ}$
$20\, \sin 40^{\circ}$
$\dfrac{20}{\cos 40^{\circ}}$
$\dfrac{20}{\sin 40^{\circ}}$

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$A$

Show Worked Solution

$\cos 40^{\circ}$	$=\dfrac{x}{20}$
$x$	$=20\, \times \cos 40^{\circ}$

$\Rightarrow A$

Measurement, STD2 M6 2011 FUR1 2

The point `Q` on building `B` is visible from the point `P` on building `A`, as shown in the diagram above.

Building `A` is 16 metres taller than building `B`.

The horizontal distance between point `P` and point `Q` is 23 metres.

Calculate the angle of depression of point `Q` from point `P`. Give your answer correct to 1 decimal place. (2 marks)

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`34.8^@`

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`tan theta`	`= 16/23`
`theta`	`= 34.8^@`

`:.\ text(Angle of depression from)\ P\ text(to)\ Q = 34.8^@`

`text{(alternate,}\ PS\ text(||)\ RQ text{)}`

Measurement, STD2 M6 2020 HSC 32

The diagram shows a regular decagon (ten-sided shape with all sides equal and all interior angles equal). The decagon has centre `O`.

The perimeter of the shape is 80 cm.

By considering triangle `OAB`, calculate the area of the ten-sided shape. Give your answer in square centimetres correct to one decimal place. (4 marks)

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`492.4\ text(cm²)`

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`angle AOB = 360/10 = 36^@`

♦♦ Mean mark 33%.

`AB = 80/10 = 8\ text(cm)`

`DeltaAOB\ text(is made up of 2 identical right-angled triangles)`

`tan 18^@`	`= 4/x`
`x`	`= 4/(tan 18^@)`

`:.\ text(Area of decagon)`	`= 20 xx 1/2 xx 4/(tan 18^@) xx 4`
	`= 492.429…`
	`= 492.4\ text(cm² (to 1 d.p.))`

Measurement, STD2 M6 2020 HSC 16

Consider the triangle shown.

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

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

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a. `39^@`

b. `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?

3.7 m
5.9 m
8.3 m
13.2 m

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`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)

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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 EQ-Bank 2 MC

In which triangle is `sin theta = 4/7`?

A.		B.
C.		D.

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`D`

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`=>D`

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)}`

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`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 2017 HSC 8 MC

The diagram shows a right-angled triangle.

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

`70°16^{′}`
` 70°17^{′}`
`70°27^{′}`
`70°28^{′}`

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`B`

Show Worked Solution

`tan theta`	`= text(opp)/text(adj)`
	`= 5.3/1.9`
`:. theta`	`=tan^(-1)(5.3/1.9)`

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 8 MC

If `tan theta = 85`, what is the value of `theta`, correct to 2 decimal places?

`1.37°`
`1.56°`
`89.33°`
`89.20°`

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`C`

Show Worked Solution

`tan theta`	`=85`
`theta`	`=tan^(-1)85=89.33°`

`=>C`

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?

60 m
74 m
90 m
100 m

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`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 2005 HSC 25b

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

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

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a. `text(Proof)`

b. `67°23^{′}`

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a. `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”.

b. `sin ∠ABC = 12/13`

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

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?

12.7 m
14.0 m
33.1 m
64.3 m

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`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 2004 HSC 5 MC

What is the correct expression for tan 20° in this triangle?

`a/b`
`a/c`
`c/b`
`c/a`

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`D`

Show Worked Solution

`tan 20^@= text(opposite)/text(adjacent)= c/a`

`=> D`

Measurement, STD2 M6 2007 HSC 8 MC

What is the length of the side `MN` in the following triangle, correct to two decimal places?

9.19 cm
10.07 cm
15.66 cm
18.67 cm

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`C`

Show Worked Solution

`sin 50^@`	`= 12/(MN)`
`MN`	`= 12/sin50^@= 15.66\ …`

`=> C`

Measurement, STD2 M6 2008 HSC 14 MC

Danni is flying a kite that is attached to a string of length 80 metres. The string makes an angle of 55° with the horizontal.

How high, to the nearest metre, is the kite above Danni’s hand?

46 m
66 m
98 m
114 m

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`B`

Show Worked Solution

`sin 55^@`	`= h/80`
`h`	`= 80 xx sin 55^@= 65.532\ …\ text(m)`

`=> B`

Measurement, STD2 M6 2014 HSC 26b

Calculate the value of `h` correct to two decimal places. (2 marks)

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`10.65\ text{(2 d.p.)}`

Show Worked Solution

`sin 28^@`	`= 5/h`
`:. h`	`= 5/(sin 28^@)`
	`= 10.6502\ …`
	`= 10.65\ \ text{(2 d.p.)}`

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.

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

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

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a. `53^circ`

b. `190\ text(m)`

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♦♦ Mean mark (a) 31%

a.	`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^circ\ text{(nearest degree)}`

b. `text(Find)\ CB:`

♦♦ Mean mark (b) 31%
MARKER’S COMMENT: The most efficient solution uses right-angled trigonometry.

`/_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°.

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

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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)

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

b. `17^@28^{′}`

Show Worked Solution

a. `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.)`

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

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

`8/cos30°`
`8/sin30°`
`8 xx cos30°`
`8 xx sin30°`

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`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)

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`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?

10 m
11 m
13 m
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?

`29 xx cos40^@`
`29 xx cos 50^@`
`cos40^@/29`
`cos50^@/29`

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`A`

Show Worked Solution

♦ Mean mark 42%

`cos40^@`	`= x/29`
`x`	`= 29xxcos40^@`

`=> A`

Measurement, STD2 M6 2013 HSC 4 MC

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

`31^@`
`37^@`
`49^@`
`53^@`

Show Answers Only

`B`

Show Worked Solution

`sin theta`	`=81/135`
`:. theta`	`=37^@`

`=>\ B`

`text(Note that)\ tan theta\ text(or)\ cos theta\ text(could also be used.)`

\(\cos 40^{\circ}\)	\(=\dfrac{x}{20}\)
\(x\)	\(=20\, \times \cos 40^{\circ}\)