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EXAMCOPY Functions, 2ADV F1 2009 HSC 1a

Sketch the graph of  `y-2x = 3`, showing the intercepts on both axes.   (2 marks)

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Show Worked Solution

`y-2x=3\ \ =>\ \ y=2x+3`

`ytext{-intercept}\ = 3`

`text{Find}\ x\ text{when}\ y=0:`

`0-2x=3\ \ =>\ \ x=-3/2`
 

EXAMCOPY Algebra, STD2 A1 2004 HSC 11 MC

If  `d = 6t^2`, what is a possible value of  `t`  when  `d = 2400`?

  1. `0.05`
  2. `20`
  3. `120`
  4. `400`
Show Answers Only

`B`

Show Worked Solution
`d` `= 6t^2`
`t^2` `= d/6`
`t` `= +- sqrt(d/6)`

 
`text(When)\ \ d = 2400:`

`t` `= +- sqrt(2400/6)`
  `= +- 20`

 
`=> B`

Special Properties, SMB-031

In the diagram, `ABCDE` is a regular pentagon. The diagonals `AC` and `BD` intersect at `F`.

  1. Show that the size of `/_ABC` is 108°.  (1 mark)

  2. Find the size of `/_BAC`. Give reasons for your answer.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `36°`
Show Worked Solution
i.    

`text(Sum of all internal angles)`

`= (n-2) xx 180°`

`= (5-2) xx 180°`

`= 540°`
 

`:. /_ABC= 540/5= 108°`
 

ii.  `BA = BC\ \ text{(equal sides of a regular pentagon)}`

`:. Delta BAC\ text(is isosceles)`

`/_BAC= 1/2 (180-108)=36^{\circ} \ \ \ text{(base angle of}\ Delta BAC text{)}`

Congruency, SMB-016

Igor was designing a shield using 10 congruent (isosceles) triangles, as shown in the diagram below.
 

How many degrees in the angle marked `x`?   (3 marks)

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`72^@`

Show Worked Solution

`text(Angles at centre of circle)\ = 360/10 = 36^@`

`text{Since triangles are isosceles:}`

`180` `= 36 + 2x`
`2x` `= 180-36`
  `= 144 `
`:. x` `= 72^@`

Special Properties, SMB-020 MC

`AB` is the diameter of a circle, centre `O`.

There are 3 triangles drawn in the lower semi-circle and the angles at the centre are all equal to `x^@`.

The three triangles are best described as:

  1. isosceles
  2. scalene
  3. right-angled
  4. equilateral
Show Answers Only

`D`

Show Worked Solution

`3x=180^{\circ}\ \=>\ x=60^{\circ} `

`AO=OC=OD=OB\ \ text{(radii of circle)}`

`=>\ text{Since angles opposite equal sides of a triangle are}`

`text(equal, all triangle angles can be found to equal 60°.)`

`:.\ text(The three triangles are equilateral.)`

`=>D`

Special Properties, SMB-019

A regular pentagon, a square and an equilateral triangle meet at a point.
 

 
What is the size of the angle `x°`?   (3 marks)

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`102°`

Show Worked Solution

`text(Sum of internal angles of pentagon)`

`= (n-2) xx 180`

`= 3 xx 180`

`= 540^@`
 

`text(Internal angle in pentagon)`

`= 540/5`

`= 108^@`
 

`:. x` `= 360-(108 + 90 + 60)`
  `= 102^@`

Special Properties, SMB-018

A six sided figure is drawn below.
  

What is the sum of the six interior angles?   (2 marks)

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`720^@`

Show Worked Solution

`\text{Method 1}`

`text(Reflex angle) = 360-90 = 270^@`

`:.\ text(Sum of interior angles)`

`= (270 xx 2) + (30 xx 2) + (60 xx 2)`

`= 720^@`
 

`\text{Method 2}`

`text{Sum of interior angles (formula)}`

`= (n-2) xx 180`

`=4 xx 180`

`= 720^@`

Special Properties, SMB-016

Two identical quadrilaterals fit together to make this regular pentagon.
 

What is the value of `x`?   (2 marks)

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`108^@`

Show Worked Solution

`text(Consider regular pentagon:)`

`text{Sum of internal angles (formula)}`

`= (n-2) xx 180`

`= 3 xx 180`

`= 540^@`
 

`:. x` `= 540/5`
  `= 108^@`
TIP: Two quadrilaterals joining does not make the internal angle sum = 2 × 360°!

Special Properties, SMB-015

A star is drawn on the inside of a regular pentagon, as shown below.
 

What is the size of the angle marked `x`?   (3 marks)

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`36^@`

Show Worked Solution

`text(Consider the triangle)\ ABC\ \ text(in the pentagon:)`

STRATEGY: The internal angle sum is  3 × 180 = 540 (since 3 triangles can be drawn internally from one point).

`text(Total degrees in a pentagon)`

`= 3 xx 180`

`= 540^@`

 
`text{Internal angle}\ =540/5 = 108^@\ \ \text{(regular pentagon)}`
  
`DeltaABC\ text(is isosceles)`

`:. x + x + 108` `= 180`
`2x` `= 72`
`x` `= 36^@`

Special Properties, SMB-014

The sum of the interior angles of a 6 sided polygon can be found by first dividing it into triangles from one vertex.
 

 

What is the sum of the interior angles of this polygon?   (2 marks)

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`720\ text(degrees)`

Show Worked Solution

`text{Since the polygon can be divided into 4 separate triangles:}`

`text(Sum of interior angles)`

`= 4 xx 180`

`= 720\ text(degrees)`

Special Properties, SMB-013 MC

Which statement is always true?

  1. Scalene triangles have two angles that are equal.
  2. All angles in a parallelogram are equal.
  3. The opposite sides of a trapezium are equal in length.
  4. The diagonals of a rhombus are perpendicular to each other.
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`D`

Show Worked Solution

`text{Consider each option:}`

`A:\ \text{Isosceles (not scalene) have two equal angles.}`

`B:\ \text{Only opposite angles in a parallelogram are equal.}`

`C:\ \text{At least one pair of opposite sides of a trapezium are not equal.}`

`D:\ \text{Rhombuses have perpendicular diagonals.}`

`=>D`

Special Properties, SMB-010 MC

Which of these are always equal in length?

  1. the diagonals of a rhombus
  2. the diagonals of a parallelogram
  3. the opposite sides of a parallelogram
  4. the opposite sides of a trapezium
Show Answers Only

`C`

Show Worked Solution

`text{Consider each option:}`

`A:\ \text{rhombus diagonals are perpendicular but not always equal}`

`B:\ \text{parallelogram diagonals not always equal (see below)}`

`C:\ \text{always true (see above)}`

`D:\ \text{at least 1 pair of opposite sides of a trapezium are not equal}`
\(\Rightarrow C\)

Special Properties, SMB-009 MC

`PQRS` is a parallelogram.

Which of these must be a property of `PQRS`?

  1. Line `PS` is perpendicular to line `PQ`.
  2. Line `PQ` is parallel to line `PS`.
  3. Diagonals `PR` and `SQ` are perpendicular.
  4. Line `PS` is parallel to line `QR`.
Show Answers Only

`D`

Show Worked Solution

`text{By elimination:}`

`A\ \text{and}\ B\ \text{clearly incorrect.}`

`C\ \text{true if all sides are equal (rhombus) but not true for all parallelograms.}`

`text(Line)\ PS\ text(must be parallel to line)\ QR.`

`=>D`

Special Properties, SMB-008

The sum of the internal angles of a polygon can be calculated by drawing triangles from any given vertex as shown below. 
 

What is the size of the angle marked `x` in the diagram below?   (2 marks)

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`107°`

Show Worked Solution

`text{Since the quadrilateral was divided into two triangles}`

`=>\ \text{Sum of internal angles}\ = 2 xx 180 = 360^{\circ}`

`:. x` `= 360-(103 + 88 + 62)`
  `= 360-253`
  `= 107°`

Congruency, SMB-013

Which two of the triangles below are congruent?   (2 marks)

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\(\text{Triangle B and Triangle C}\)

Show Worked Solution

\(\text{Unknown side}\ (x)\ \text{in Triangle B (Pythagoras):}\)

\(x=\sqrt{11^2-(\sqrt{96})^2} = \sqrt{25} = 5\)

\(\Rightarrow\ \text{Triangle B and Triangle C are congruent (SSS)} \)

\(\text{Triangle A can be shown to have different dimensions but this is not necessary.}\)

Congruency, SMB-012

Which two of the triangles below are congruent?   (2 marks)

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\(\text{Triangle A and Triangle C}\)

Show Worked Solution

\(\text{Unknown side}\ (x)\ \text{in Triangle A (Pythagoras):}\)

\(x=\sqrt{8^2-(\sqrt{48})^2} = \sqrt{16} = 4\)

\(\Rightarrow\ \text{Triangle A and Triangle C are congruent (SSS)} \)

\(\text{Triangle B can be shown to have different dimensions but this is not necessary.}\)

Congruency, SMB-015

The two triangles below are congruent.

  1. Which congruency test would be used to prove the two triangles above are congruent?   (1 mark)

  2. Find the values of  `a` and `b`.  (2 marks)

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  1. `text{(SAS)}`
  2. `a=12.1, \ b=7.4`
Show Worked Solution

i.    `text{Congruency test}\ => \ text{SAS} `
 

ii.   `text{Matching corresponding sides:}`

`a=12.1, \ b=7.4`

Congruence, SMB-014

The area of the rectangle in the diagram below is 15 cm2.
 

Giving reasons, find the area of the trapezium `ABCD`.   (4 marks)

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`30\ text(cm)^2`

Show Worked Solution

`text{Opposite sides of rectangles are equal and parallel}`

`\Delta AEI ≡ \Delta DHJ\ \ text{(RHS)}`

`/_ EAI = /_ BEF\ \ text{(corresponding,}\ AD\ text{||}\ EH )`

`\Delta AEI ≡ \Delta EBF\ \ text{(AAS)}`

`text{Similarly,}\ \Delta DHJ ≡ \Delta HCG\ \ text{(AAS)}`

`=>\ \text{All triangles in diagram are congruent.}`

`text(Rearranging the diagram:)`

`:.\ text(Area of trapezium)`

`= 2 xx 15`

`= 30\ text(cm)^2`

Special Properties, SMB-004 MC

Which one of the following triangles is impossible to draw?

  1. a right angled triangle with two acute angles
  2. an isosceles triangle with one right angle
  3. a scalene triangle with three acute angles
  4. a right angled triangle with one obtuse angle
Show Answers Only

`D`

Show Worked Solution

`text(A right angle = 90°.)`

`text{Since an obtuse angle is greater than 90°, it is impossible for}`

`text(a triangle, with an angle sum less than 180°, to have both.)`

`=>D`

Special Properties, SMB-003 MC

Select the statement that is true about triangle `ABC`.

  1. Triangle `ABC` is a scalene triangle.
  2. Triangle `ABC` has exactly 2 equal sides.
  3. Triangle `ABC` is an equilateral triangle.
  4. Triangle `ABC` is an obtuse triangle.
Show Answers Only

`C`

Show Worked Solution

`angleA = 180-(60 + 60) = 60^@`

`:. text(All angles are)\ 60^@.`

`:. text(Triangle)\ ABC\ text(is an equilateral.)`

`=>C`

Special Properties, SMB-002 MC

A triangle has two acute angles.

What type of angle couldn't the third angle be?

  1. an acute angle
  2. an obtuse angle
  3. a right-angle
  4. a reflex angle
Show Answers Only

`D`

Show Worked Solution

`text(A triangle’s angles add up to 180°, and a reflex angle is)`

`text(greater than 180°.)`

`:.\ text(The third angle cannot be reflex.)`

`=>D`

Special Properties, SMB-001 MC

Which of the following triangle types is impossible to draw?

  1. a right-angled, scalene triangle
  2. a right-angled, equilateral triangle
  3. an obtuse-angled, isosceles triangle
  4. an acute-angled, scalene triangle
Show Answers Only

`B`

Show Worked Solution

`text(An equilateral triangle has all angles = 60°.)`

`:.\ text(A right-angled, equilateral triangle is impossible.)`

`=>B`

Congruency, SMB-011

The diagram shows a circle with centre `O` and radius 5 cm.

The length of the arc `PQ` is 9 cm. Lines drawn perpendicular to `OP` and `OQ` at `P` and `Q` respectively meet at  `T`.
 

Prove that `Delta OPT` is congruent to `Delta OQT`.   (2 marks)

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

Show Worked Solution

`text(Prove)\ Delta OPT ≡ Delta OQT`

`OT\ text(is common)`

COMMENT: Know the difference between the congruency proof of `RHS` and `SAS`. Incorrect identification will lose a mark.

`/_OPT = /_OQT = 90°\ \ text{(given)}`

`OP = OQ\ \ \ text{(radii)}`

`:.\ Delta OPT ≡ Delta OQT\ \ \ text{(RHS)}`

Congruency, SMB-010

In the diagram, `AD` is parallel to `BC`, `AC` bisects `/_BAD` and `BD` bisects `/_ABC`. The lines `AC` and `BD` intersect at `P`.

  1. Prove that `/_BAC = /_BCA`.  (1 mark)

  2. Prove that `Delta ABP ≡ Delta CBP`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.  

`text(Prove)\ /_BAC = /_BCA`

`/_BCA` `= /_CAD\ \ \ text{(alternate angles,}\ BC\ text(||)\ AD text{)}`
`/_CAD` `= /_BAC\ \ \ text{(}AC\ text(bisects)\ /_BAD text{)}`
`:. /_BAC` `= /_BCA\ …\ text(as required)`

 

ii.  `text(Prove)\ Delta ABP ≡ Delta CBP`

`/_BAC` `= /_BCA\ \ \ text{(from part (i))}`
`/_ABP` `= /_CBP\ text{(}BD\ text(bisects)\ /_ABC text{)}`
`BP\ text(is common)`

 

`:. Delta ABP ≡ Delta CBP\ \ text{(AAS)}`

Congruency, SMB-008

In the figure below, \(ABCD\) is a parallelogram where opposite sides of the quadrilateral are equal.
 

Prove that a diagonal of the parallelogram produces two triangles that are congruent.  (2 marks)

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\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(\text{One of multiple solutions:}\)

\( AB=CD\ \ \text{and}\ \ AC=BD\ \ \text{(given)} \)

\(BC\ \text{is common} \) 

\(\therefore\ \Delta ABC \equiv \Delta DCB\ \ \text{(SSS)}\)

Congruency, SMB-007

In the figure below, \(AB \parallel DE, \ AC = CE\)  and the line \(AE\) intersects \(DB\) at \(C\).
 

Prove that this pair of triangles are congruent.  (2 marks)

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\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\( \angle ACB = \angle DCE\ \ \text{(vertically opposite)} \)

\(AC = CE\ \ \text{(given)} \)

\(\angle BAC = \angle DEC\ \ \text{(alternate,}\ AB \parallel DE \text{)} \)
 

\(\therefore\ \Delta ABC \equiv \Delta EDC\ \ \text{(AAS)}\)

Congruency, SMB-006

In the quadrilateral \(ABCD\), \(AB \parallel CD, \angle BAD = \angle BCD\)  and  \(\angle DBC = \angle BDA = 90^{\circ} \).
 

Prove that this pair of triangles are congruent.  (2 marks)

Show Answers Only

\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(\angle BAD = \angle BCD\ \ \text{(given)} \) 

\(\angle DBC = \angle BDA = 90^{\circ} \ \ \text{(given)} \)

\(BD\ \text{is common} \) 
 

\(\therefore\ \Delta BAD \equiv \Delta DCB\ \ \text{(AAS)}\)

Congruency, SMB-005

In the figure below, \(BE = BC\), \(AB = BD\) and the line \(AD\) intersects \(CE\) at \(B\).
 

Prove that this pair of triangles are congruent.  (2 marks)

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\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\( \angle ABE = \angle CBD\ \ \text{(vertically opposite)} \)

\(AB = BD\ \ \text{(given)} \)

\(EB = BC\ \ \text{(given)} \)
 

\(\therefore\ \Delta ABE \equiv \Delta DBC\ \ \text{(SAS)}\)

Congruency, SMB-004

In the circle below, centre \(O\), \(OB\) is perpendicular to chord \(AC\).
 

Prove that a pair of triangles in this figure are congruent.  (2 marks)

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\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(\text{One of multiple proofs:}\)

\(OB\ \ \text{common} \)

\( \angle OBA = \angle OBC = 90^{\circ}\ \ \text{(given)} \)

\(OA = OC\ \ \text{(radii)} \)
 

\(\therefore\ \Delta AOB \equiv \Delta COB\ \ \text{(RHS)}\)

Congruency, SMB-003

In the figure below, the line \(AD\) intersects \(BE\) at \(C\), \(BC = CD\) and \(AC = EC\).
 

Prove that this pair of triangles are congruent.  (2 marks)

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\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\( \angle BCE = \angle DCE\ \ \text{(vertically opposite)} \)

\(BC = CD\ \ \text{(given)} \)

\(AC = EC\ \ \text{(given)} \)
 

\(\therefore\ \Delta ABC \equiv \Delta EDC\ \ \text{(SAS)}\)

Congruency, SMB-002

The diagram shows two triangles that touch at the middle of a circle.

Prove that this pair of triangles are congruent.  (2 marks)

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\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(\text{Base of each triangle (chords) are equal} \)

\(\text{All other sides are equal radii of the circle} \)

\(\therefore\ \text{Two given triangles are congruent}\ \ \text{(SSS)}\)

Congruency, SMB-001

The diagram shows two right-angled triangles where \(\angle BAC = \angle BDC =  90^{\circ}\), and  \(AB = BD\). 

Prove that this pair of triangles are congruent.  (2 marks)

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\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(BC\ \text{(hypotenuse) is common} \)

\(BA = BD\ \text{(given)} \)

\(\angle BAC = \angle BDC =  90^{\circ}\ \ \text{(given)} \)

\(\therefore \Delta ABC \equiv \Delta DBC\ \ \text{(RHS)}\)

Similarity, SMB-026

The two triangles below are similar.
 

Find the length of \(ED\).   (3 marks)

Show Answers Only

\(ED = 15\)

Show Worked Solution

\(\text{Using Pythagoras in}\ \Delta ABC: \)

\( AB=\sqrt{13^2-12^2}=\sqrt{25}=5 \)

\(\text{Scale factor}\ = \dfrac{FD}{AC} = \dfrac{39}{13} = 3 \)

\(\therefore ED = 3 \times AB = 3 \times 5 = 15 \)

Similarity, SMB-024

Prove that the two triangles in the right cone pictured below are similar.   (2 marks)
 

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\(\text{Proof (See Worked Solution)}\)

Show Worked Solution

\(\angle ACD \ \text{is common to}\ \Delta ACE\ \text{and}\ \Delta BCD \)

\(\angle CAE=\angle CBD=90^\circ \ \text{(Right cone)} \)

\(\therefore \Delta ACE\ \text{|||}\ \Delta BCD\ \ \text{(equiangular)}\)

Similarity, SMB-023

Show that \(\Delta ACD\) and \(\Delta DCB\) in the figure below are similar.   (2 marks)
 

Show Answers Only

\(\text{Proof (See Worked Solution)}\)

Show Worked Solution

\(\angle BDC = 180-(88+44) = 48^{\circ} \ \text{(angle sum of Δ)} \)

\(\angle CAD = \angle CDB = 48^{\circ} \)

\(\angle ACD=44^{\circ} \ \text{is common to}\ \Delta ACD\ \text{and}\ \Delta BCD \)

\(\therefore \Delta ACD\ \text{|||}\ \Delta DCB\ \ \text{(equiangular)}\)

Similarity, SMB-022

Prove that this pair of triangles are similar.   (2 marks)

Show Answers Only

\(\text{Proof (See Worked Solution)}\)

Show Worked Solution

\(\angle PQT = \angle SQR \ \text{(vertically opposite)} \)

\(\dfrac{PQ}{QS} = \dfrac{2.5}{5} = \dfrac{1}{2} \)

\(\dfrac{TQ}{QR} = \dfrac{2}{4} = \dfrac{1}{2} \)

\(\therefore \Delta PQT\ \text{|||}\ \Delta SQR\ \ \text{(sides adjacent to equal angles in proportion)}\)

Similarity, SMB-021

Prove that this pair of triangles are similar.   (2 marks)
 

Show Answers Only

\(\text{Proof (See Worked Solution)}\)

Show Worked Solution

\(\angle ACD = \angle BEA \ \text{(given)} \)

\(\angle BAE\ \text{is common to}\ \Delta ACD\ \text{and}\ \Delta BEA \)

\(\therefore \Delta ACD\ \text{|||}\ \Delta BEA\ \ \text{(equiangular)}\)

Similarity, SMB-020

Prove that this pair of triangles are similar.   (3 marks)
 

Show Answers Only

\(\text{Proof (See Worked Solution)}\)

Show Worked Solution

\(\text{Find unknown side}\ (x)\ \text{of smaller triangle}\)

\(\text{Using Pythagoras:}\)

\(x=\sqrt{5^2-4^2} = \sqrt{9} = 3\)
 

\(\dfrac{AC}{DE} = \dfrac{5}{15} = \dfrac{1}{3} \)

\(\dfrac{BC}{EF} = \dfrac{3}{9} = \dfrac{1}{3} \)

\(\angle ABC = \angle DEF = 90^{\circ} \ \ \text{(given)} \)

\(\therefore \Delta ABC\ \text{|||}\ \Delta DEF\ \ \text{(hypotenuse and second side of right-angled triangle in proportion)}\)

Similarity, SMB-019

Prove that this pair of triangles are similar.   (2 marks)
 

Show Answers Only

\(\text{Proof (See Worked Solution)}\)

Show Worked Solution

\(\dfrac{AB}{GH} = \dfrac{11}{22} = \dfrac{1}{2} \)

\(\dfrac{BC}{FG} = \dfrac{3.5}{7} = \dfrac{1}{2} \)

\(\angle ABC = \angle FGH = 95^{\circ} \ \ \text{(given)} \)
 

\(\therefore \Delta ABC\ \text{|||}\ \Delta HGF\ \ \text{(sides adjacent to equal angles in proportion)}\)

Similarity, SMB-018

Prove that this pair of triangles are similar.   (2 marks)
 

Show Answers Only

\(\text{Proof (See Worked Solution)}\)

Show Worked Solution

\(\angle ABE = \angle CBD\ \ \text{(vertically opposite)} \)

\(\angle AEB = \angle BCD\ \ \text{(alternate,}\ AE \parallel CD \text{)} \)

\(\therefore \Delta ABE\ \text{|||}\ \Delta DBC\ \ \text{(equiangular)}\)

Similarity, SMB-017

Prove that this pair of triangles are similar.   (2 marks)
 

Show Answers Only

\(\text{Proof (See Worked Solution)}\)

Show Worked Solution

\(\dfrac{RP}{AC} = \dfrac{18}{12} = \dfrac{3}{2} \)

\(\dfrac{QR}{BA} = \dfrac{12}{8} = \dfrac{3}{2} \)

\(\dfrac{PQ}{CB} = \dfrac{9}{6} = \dfrac{3}{2} \)
 

\(\therefore \Delta PQR\ \text{|||}\ \Delta CBA\ \ \text{(three pairs of sides in proportion)}\)

Congruency, SMB-009

 

The diagram shows a right-angled triangle `ABC` with `∠ABC = 90^@`. The point `M` is the midpoint of `AC`, and `Y` is the point where the perpendicular to `AC` at `M` meets `BC`.

Show that `\Delta AYM \equiv \Delta CYM`.  (2 marks)

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

Show Worked Solution

`text(In)\ ΔAYM\ text(and)\ ΔCYM`

`∠AMY` `= ∠CMY = 90^@\ \ \ (MY ⊥ AC)`
`AM` `=CM\ \ \ text{(given)}`
`YM\ text(is common)`

 
`:. \Delta AYM \equiv \Delta CYM\ \ text{(SAS)}`

Trigonometry, SMB-067

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

Show Answers Only

\(\theta=112^{\circ} 12^{′}\)

Show Worked Solution
\(\cos A\) \(= \dfrac{b^2+c^2-a^2}{2bc} \)  
\(\cos \theta\) \(= \dfrac{7.8^2+10.2^2-15.0^2}{2 \times 7.8 \times 10.2}\)  
  \(= -0.3778…\)  
\(\therefore \theta\) \(= \cos^{-1} (-0.3778…) \)  
  \(=112.199…^{\circ} \)  
  \(=112^{\circ} 12^{′}\ \ \text{(nearest minute)} \)  

Trigonometry, SMB-066

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

Show Answers Only

\(\theta=35^{\circ}\)

Show Worked Solution
\(\cos A\) \(= \dfrac{b^2+c^2-a^2}{2bc} \)  
\(\cos \theta\) \(= \dfrac{16.2^2+18.1^2-10.5^2}{2 \times 16.2 \times 18.1}\)  
  \(= 0.818…\)  
\(\therefore \theta\) \(= \cos^{-1} (0.818…) \)  
  \(=35.09…^{\circ} \)  
  \(=35^{\circ}\ \ \text{(nearest degree)} \)  

Trigonometry, SMB-065

Find the value of \(\theta\), correct to 1 decimal place.   (2 marks)

Show Answers Only

\(\theta=38.9^{\circ}\)

Show Worked Solution
\(\cos A\) \(= \dfrac{b^2+c^2-a^2}{2bc} \)  
\(\cos \theta\) \(= \dfrac{9^2+5^2-6^2}{2 \times 9 \times 5}\)  
  \(= 0.777…\)  
\(\therefore \theta\) \(= \cos^{-1} (0.777…) \)  
  \(=38.94…^{\circ} \)  
  \(=38.9^{\circ}\ \ \text{(1 d.p.)} \)  

Trigonometry, SMB-064

Find the value of \(d\), correct to 2 decimal places.   (3 marks)

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\(d=12.9\ \text{km}\)

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\(a^2\) \(=b^2+c^2-2bc\ \cos A \)  
\(d^2\) \(= 8.9^2+5.2^2-2 \times 8.9 \times 5.2 \times \cos 130^{\circ}\)  
  \(= 165.746…\)  
\(\therefore d\) \(=12.87…\)  
  \(=12.9\ \text{km (1 d.p.)} \)  

Trigonometry, SMB-063

Find the value of \(x\), correct to 1 decimal place.   (3 marks)
 

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\(x=15.4\ \text{cm}\)

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\(a^2\) \(=b^2+c^2-2bc\ \cos A \)  
\(x^2\) \(= 15^2+25^2-2 \times 15 \times 25 \times \cos 35^{\circ}\)  
  \(= 235.63…\)  
\(\therefore x\) \(=15.35…\)  
  \(=15.4\ \text{cm (1 d.p.)} \)  

Trigonometry, SMB-062

Find the value of \(x\), correct to 1 decimal place.   (3 marks)
 

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\(x=8.8\ \text{m}\)

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\(a^2\) \(=b^2+c^2-2bc\ \cos A \)  
\(x^2\) \(= 5^2+9^2-2 \times 5 \times 9 \times \cos 72^{\circ}\)  
  \(= 78.188…\)  
\(\therefore x\) \(=8.842…\)  
  \(=8.8\ \text{m (1 d.p.)} \)  

Trigonometry, SMB-061

Find the value of \(\theta\), correct to the nearest degree.   (3 marks)

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\(\theta=154^{\circ}\)

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\(\dfrac{\sin \theta}{21}\) \(= \dfrac{\sin 12^{\circ}}{10} \)  
\(\sin \theta\) \(= \dfrac{21 \times \sin 12^{\circ}}{10}\)  
\(\theta\) \(=\sin^{-1} (0.4366) \)  
  \(=25.88…^{\circ} \)  
  \(=26^{\circ}\ \text{(nearest degree)} \)  

 
\(\therefore\ \theta = 180-26=154^{\circ}\ \ \text{(angle is obtuse)}\)

Trigonometry, SMB-060

Find the value of \(\alpha\), correct to the nearest degree.   (2 marks)

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\(\alpha=39^{\circ}\)

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\(\dfrac{\sin \alpha}{9}\) \(= \dfrac{\sin 79^{\circ}}{14} \)  
\(\sin \alpha\) \(= \dfrac{9 \times \sin 79^{\circ}}{14}\)  
\(\alpha\) \(=\sin^{-1} (0.6310) \)  
  \(=39.12…^{\circ} \)  
  \(=39^{\circ}\ \text{(nearest degree)} \)  

Trigonometry, SMB-059

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

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\(\theta=50^{\circ}\)

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\(\dfrac{\sin \theta}{7.5}\) \(= \dfrac{\sin 66^{\circ}}{8.9} \)  
\(\sin \theta\) \(= \dfrac{7.5 \times \sin 66^{\circ}}{8.9}\)  
\(\theta\) \(=\sin^{-1} (0.7698) \)  
  \(=50.33…^{\circ} \)  
  \(=50^{\circ}\ \text{(nearest degree)} \)  

Trigonometry, SMB-058

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

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\(\theta=53^{\circ}\)

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\(\dfrac{\sin \theta}{6}\) \(= \dfrac{\sin 37^{\circ}}{4.5} \)  
\(\sin \theta\) \(= \dfrac{6 \times \sin 37^{\circ}}{4.5}\)  
\(\theta\) \(=\sin^{-1} (0.8024) \)  
  \(=53.361…^{\circ} \)  
  \(=53^{\circ}\ \text{(nearest degree)} \)  

Trigonometry, SMB-057

Find \(\alpha\), to the nearest degree, such that

\(\dfrac{\sin \alpha}{8} = \dfrac{\sin 60^{\circ}}{11} \)   (2 marks)

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\(\alpha=39^{\circ}\)

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\(\dfrac{\sin \alpha}{8}\) \(= \dfrac{\sin 60^{\circ}}{11} \)  
\(\sin \alpha\) \(= \dfrac{8 \times \sin 60^{\circ}}{11}\)  
\(\alpha\) \(=\sin^{-1} (0.6298) \)  
  \(=39^{\circ}\ \text{(nearest degree)} \)