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Trigonometry, 2ADV T1 EQ-Bank 20

A regular hexagon with centre \(O\) is divided into six identical equilateral triangles. One such triangle is shown on the diagram.
 

The area of the hexagon is \(12 \sqrt{3}\text{ cm}^2\).

By considering the area of triangle \(O A B\), calculate the exact perimeter of the hexagon.   (3 marks)

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\(\text{Since there are 6 identical triangles:}\)

\(\text{Area of }\triangle O A B=\dfrac{12 \sqrt{3}}{6}=2 \sqrt{3}\text{ cm}^2\)

\(\text{Using the sine rule (area)}:\)

\(A=\dfrac{1}{2} \times O A \times O B \times \sin 60^{\circ}=\dfrac{\sqrt{3}}{4} x^2\)

\(\text{Equating areas:}\)

\(\dfrac{\sqrt{3}}{4} x^2\) \(=2\sqrt{3}\)
\(x^2\) \(=8\)
\(x\) \(=\sqrt{8}=2\sqrt{2}\)

 

\(A B=2 \sqrt{2} \ \ \text{(since} \ \ \triangle O A B \ \ \text{is equilateral)}\)

\(\therefore\ \text {Perimeter of hexagon}=6 \times 2 \sqrt{2}=12 \sqrt{2}\text{ cm}\)

Show Worked Solution

\(\text{Since there are 6 identical triangles:}\)

\(\text{Area of }\triangle O A B=\dfrac{12 \sqrt{3}}{6}=2 \sqrt{3}\text{ cm}^2\)

\(\text{Using the sine rule (area)}:\)

\(A=\dfrac{1}{2} \times O A \times O B \times \sin 60^{\circ}=\dfrac{\sqrt{3}}{4} x^2\)

\(\text{Equating areas:}\)

\(\dfrac{\sqrt{3}}{4} x^2\) \(=2\sqrt{3}\)
\(x^2\) \(=8\)
\(x\) \(=\sqrt{8}=2\sqrt{2}\)

 

\(A B=2 \sqrt{2} \ \ \text{(since} \ \ \triangle O A B \ \ \text{is equilateral)}\)

\(\therefore\ \text {Perimeter of hexagon}=6 \times 2 \sqrt{2}=12 \sqrt{2}\text{ cm}\)

Trigonometry, 2ADV T1 EQ-Bank 6 MC

The triangle \(A B C\) has an exact area of \((12-\sqrt{3})\) cm\(^{2}\).
 

The exact value of \(x\) is

  1. \(\dfrac{24-2\sqrt{3}}{3}\)
  2. \(\dfrac{8\sqrt{3}-2}{3} \)
  3. \(8\sqrt{3}-1\)
  4. \(12\sqrt{3}+1\)
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\(B\)

Show Worked Solution
  \(12-\sqrt{3}\) \(=\dfrac{1}{2} \cdot 6 \cdot x \cdot \sin 120^{\circ}\)
  \(12-\sqrt{3}\) \(=3x \cdot \sin 120^{\circ}\)
  \(12-\sqrt{3}\) \(=\dfrac{3\sqrt{3}x}{2}\)
  \(x\) \(=\dfrac{2(12-\sqrt{3})}{3\sqrt{3}} \times \dfrac {\sqrt{3}}{\sqrt{3}}\)
    \(=\dfrac{24\sqrt{3}-6}{9} \)
    \(=\dfrac{8\sqrt{3}-2}{3} \)

 
\(\Rightarrow B\)

Trigonometry, 2ADV T1 2018 HSC 14a

In  `Delta KLM, KL`  has length 3, `LM` has length 6 and `/_KLM` is 60°. The point `N` is chosen on side `KM` so that `LN` bisects `/_KLM`. The length `LN` is `x`.
 


 

  1. Find the exact value of the area of `Delta KLM`.   (1 mark)

  2. Hence, or otherwise, find the exact value of `x`.   (2 marks)

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  1. `(9 sqrt 3)/2`
  2. `2 sqrt 3`
Show Worked Solution

i.   `text(Using sine rule:)`

`text(Area)\ \ Delta KLM= 1/2 xx 3 xx 6 xx sin 60^@= (9 sqrt 3)/2\ \ text(u)^2`
 

ii.  `text(Area)\ \ Delta KLN + text(Area)\ \ Delta NLM = text(Area)\ \ Delta KLM`

`1/2 xx 3 xx x xx sin 30^@ + 1/2 xx x xx 6 xx sin 30^@ = (9 sqrt 3)/2`

♦ Mean mark 37%.

`3/4 x + 3/2 x` `= (9 sqrt 3)/2`
`9/4 x` `= (9 sqrt 3)/2`
`:. x` `= (9 sqrt 3)/2 xx 4/9`
  `= 2 sqrt 3`

Trigonometry, 2ADV T1 2015 HSC 13a

The diagram shows `Delta ABC` with sides  `AB = 6` cm, `BC = 4` cm  and  `AC = 8` cm.
 

  1. Show that  `cos A = 7/8`.   (1 mark)

  2. By finding the exact value of `sin A`, determine the exact value of the area of `Delta ABC`.   (2 marks)

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

b.    `3 sqrt 15\ \ text(cm)^2`

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a.    `text(Show)\ cos A = 7/8`

`text(Using the cosine rule:)`

`cos A` `= (b^2 + c^2-a^2)/(2bc)`
  `= (8^2 + 6^2-4^2)/(2 xx 8 xx 6)`
  `= (64 + 36-16)/96`
  `= 84/96`
  `= 7/8\ \ text(…  as required)`

 

♦ Mean mark 40%.
b.    2UA HSC 2015 13ai
`a^2 + 7^2` `= 8^2`
`a^2 + 49` `= 64`
`a^2` `= 15`
`a` `= sqrt 15`
`:.\ sin A` `= (sqrt 15)/8`

 

`:.\ text(Area)\ Delta ABC` `= 1/2 bc\ sin A`
  `= 1/2 xx 8 xx 6 xx (sqrt 15)/8`
  `= 3 sqrt 15\ \ text(cm²)`

Trigonometry, 2ADV T1 2006 HSC 4a

In the diagram, `ABCD` represents a garden. The sector  `BCD`  has centre `B` and  `/_DBC = (5 pi)/6`

The points `A, B` and `C` lie on a straight line and  `AB = AD = 3` metres.

Copy or trace the diagram into your writing booklet.

  1. Show that `/_DAB = (2 pi)/3.`   (1 mark)

  2. Find the length of `BD`.   (2 marks)

  3. Find the area of the garden `ABCD`.   (2 marks)

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

b.    `3 sqrt 3\ \ text(m)`

c.    `(9 sqrt 3 + 45 pi) / 4\ \ text(m)^2`

Show Worked Solution
a.    

`text(Show)\ /_DAB = (2 pi)/3:`

`/_DBA= pi-(5 pi)/6= pi/6\ text{(π radians in straight angle}\ ABC text{)}`

`/_BDA = pi/6\ text(radians)\ \ text{(base angles of isosceles}\ Delta ADB text{)}`

`:. /_DAB= pi-(pi/6 + pi/6)\ \ text{(angle sum of}\ Delta ADB text{)}= (2 pi)/3`
 

b.    `text(Using the cosine rule:)`

`BD^2` `= AD^2 + AB^2-2 xx AD xx AB xx cos {:(2 pi)/3`
  `= 9 + 9-(2 xx 3 xx 3 xx -0.5)= 27`
  `= 27`
`:. BD` `= sqrt 27= 3 sqrt 3\ \ text(m)`

 

c.    `text(Area of)\ Delta ADB` `= 1/2 ab sin C`
  `= 1/2 xx 3 xx 3 xx sin{:(2 pi)/3`
  `= 9/2 xx sqrt3/2`
  `= (9 sqrt 3)/4\ \ text(m²)`

 
`text(Area of sector)\ BCD`

`= {(5 pi)/6}/(2 pi) xx pi r^2`

`= (5 pi)/12 xx (3 sqrt 3)^2`

`= (45 pi)/4\ \ text(m)^2`

 

`:.\ text(Area of garden)\ ABCD`

`= (9 sqrt 3)/4 + (45 pi)/4`

`= (9 sqrt 3 + 45 pi)/4\ \ text(m)^2`

Trigonometry, 2ADV T1 2005 HSC 3b

The lengths of the sides of a triangle are 7 cm, 8 cm and 13 cm.

  1. Find the size of the angle opposite the longest side.   (2 marks)

  2. Find the area of the triangle.   (1 marks)

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a.    `120^@`

b.    `14sqrt3\ text(cm)`

Show Worked Solution

a.

 Trig Ratios, 2UA 2005 HSC 3b Answer1 

`∠ABC\ \ text(is opposite the longest side.)`

`text(Using the cosine rule:)`

`cos angleABC= (7^2 + 8^2-13^2)/(2 xx 7 xx 8)=-1/2`

 `text{Since cos}\ 60^@ = 1/2\ text(and cos is negative in 2nd quadrant:)`

`∠ABC= 180-60= 120^@`
 

b.  `text(Using the sine rule:)`

`text(Area)\ ΔABC` `= 1/2 ab sin C`
  `= 1/2 xx 7 xx 8 sin 120^@`
  `= 28 xx sqrt3/2`
  `= 14sqrt3\ text(cm)^2`

Trigonometry, 2ADV T1 2009 HSC 5c

The diagram shows a circle with centre `O` and radius 2 centimetres. The points `A` and `B` lie on the circumference of the circle and  `/_AOB = theta`.
 

2009 5c  

  1. There are two possible values of `theta` for which the area of `Delta AOB` is `sqrt 3` square centimetres. One value is `pi/3`.Find the other value.    (2 marks)

  2. Suppose that  `theta = pi/3`.
  3. i.  Find the area of sector `AOB`   (1 mark)

  4. ii. Find the exact length of the perimeter of the minor segment bounded by the chord `AB` and the arc `AB`.   (2 marks)

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a.    `(2pi)/3`

b.i   `(2pi)/3\ \ text(cm)^2`

b.ii  `(2 + (2pi)/3)\ text(cm)`

Show Worked Solution

a.    `text(Area)\ Delta AOB= 1/2 ab sin theta= 1/2 xx 2 xx 2 xx sin theta= 2 sin theta`

`2 sin theta` `= sqrt 3\ \ \ text{(given)}`
`sin theta` `= sqrt3/2`
`theta` `= pi/3,\ pi-pi/3= pi/3,\ (2pi)/3`

  
`:.\ text(The other value of)\ theta\ text(is)\ (2pi)/3.`  
 

b.i   `text(Area of sector)\ AOB`

`= pi r^2 xx theta/(2pi)= 1/2 r^2 theta= 1/2 xx 2^2 xx pi/3= (2pi)/3\ text(cm)^2`
 

b.ii  `text(Using the cosine rule:)`

`AB^2` `= OA^2 + OB^2-2 xx OA xx OB xx cos theta`
  `= 2^2 + 2^2-2 xx 2 xx 2 xx cos (pi/3)`
  `= 4`
`AB` `= 2`

 
`text(Arc)\ AB= 2 pi r xx theta/(2pi)= r theta= (2pi)/3\ text(cm)`

`:.\ text(Perimeter) = (2 + (2pi)/3)\ text(cm)`