If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Solved Examples on Triangles

Samuel Dominic Chukwuemeka (SamDom For Peace) Verify your answers with these Calculators as applicable.

Pre-requisites:
(1.) Plane Geometry: Points, Lines, and Angles
(2.) Triangles
(3.) Mensuration Formulas

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE and Malta Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
State the reasons for every step
Show all work

(1.) ACT In $\triangle ABC$, the sum of the measures of $\angle A$ and $\angle B$ is $57^\circ$.
What is the measure of $\angle C$?


$ \angle A + \angle B + \angle C = 180^\circ ...Sum\:\: of\:\: \angle s \:\:of \:\:a\:\: triangle \\[3ex] \angle A + \angle B = 57^\circ \\[3ex] \implies 57 + \angle C = 180 \\[3ex] \angle C = 180 - 57 \\[3ex] \angle C = 123^\circ $
(2.) ACT Which of the following sets of $3$ lengths, in decimeters, are the side lengths of an obtuse triangle?
(Note: An obtuse triangle has $1$ angle whose measure is greater than $90^\circ$ and less than $180^\circ$.)

$ F.\:\: \{4, 4, 5\} \\[3ex] G.\:\: \{5, 12, 13\} \\[3ex] H.\:\: \{6, 8, 10\} \\[3ex] J.\:\: \{7, 10, 12\} \\[3ex] K.\:\: \{8, 11, 16\} \\[3ex] $

Let us find the squares of the sides of the triangle.
A triangle is obtuse if: $long^2 \gt side1^2 + side2^2$ ...Triangle Theorem Number (8.)

$ \underline{Test} \\[3ex] F.\:\: \{4, 4, 5\} \\[3ex] 4^2 = 16 \\[3ex] 4^2 = 16 \\[3ex] 5^2 = 25 \\[3ex] 25 \lt (16 + 16) \\[3ex] 25 \lt 32 ...Acute\:\: Triangle ...NO \\[3ex] G.\:\: \{5, 12, 13\} \\[3ex] 5^2 = 25 \\[3ex] 12^2 = 144 \\[3ex] 13^2 = 169 \\[3ex] 169 = 25 + 144 \\[3ex] 169 = 169 ...Right\:\: Triangle ...NO \\[3ex] H.\:\: \{6, 8, 10\} \\[3ex] 6^2 = 36 \\[3ex] 8^2 = 64 \\[3ex] 10^2 = 100 \\[3ex] 100 = 36 + 64 \\[3ex] 100 = 100 ...Right\:\: Triangle ...NO \\[3ex] J.\:\: \{7, 10, 12\} \\[3ex] 7^2 = 49 \\[3ex] 10^2 = 100 \\[3ex] 12^2 = 144 \\[3ex] 144 \lt (49 + 100) \\[3ex] 144 \lt 149 ...Acute\:\: Triangle ...NO \\[3ex] K.\:\: \{8, 11, 16\} \\[3ex] 8^2 = 64 \\[3ex] 11^2 = 121 \\[3ex] 16^2 = 256 \\[3ex] 256 \gt (64 + 121) \\[3ex] 256 \gt 185 ...Obtuse\:\: Triangle ...YES $
(3.) ACT In $\triangle ABC$ shown below, $m\angle A = x^\circ$, $m\angle B = (2x)^\circ$, and $m\angle C = (3x)^\circ$, AB = c inches, AC = b inches, and BC = a inches.
Which of the following inequalities correctly relates the side lengths of $\triangle ABC$?
(Note: $m\angle A$ denotes the measure of $\angle A$, and AB denotes the length of $\overline{AB}$
The triangle is NOT drawn to scale.)
Number 3

$ A.\;\; a \lt b \lt c \\[3ex] B.\;\; a \lt c \lt b \\[3ex] C.\;\; b \lt a \lt c \\[3ex] D.\;\; c \lt a \lt b \\[3ex] E.\;\; c \lt b \lt a \\[3ex] $

Triangle Theorem Number (10.)
The measure of an angle is proportional to the length of the side; side length is proportional to angle measure.
Because:

$ x \lt 2x \lt 3x: \\[3ex] a \lt b \lt c $
(4.) ACT The measure of the vertex angle of an isosceles triangle is $(x - 20)^\circ$.
The base angles each measure $(2x + 30)^\circ$.
What is the measure in degrees of one of the base angles?

$ A.\:\: 8^\circ \\[3ex] B.\:\: 28^\circ \\[3ex] C.\:\: 42\dfrac{1}{2}^\circ \\[5ex] D.\:\: 47\dfrac{1}{2}^\circ \\[5ex] E.\:\: 86^\circ \\[3ex] $

$ (x - 20) + (2x + 30) + (2x + 30) = 180 ...sum \:\:of\:\: \angle s \:\:of\:\: a\:\: \triangle \\[3ex] x - 20 + 2x + 30 + 2x + 30 = 180 \\[3ex] 5x + 40 = 180 \\[3ex] 5x = 180 - 40 \\[3ex] 5x = 140 \\[3ex] x = \dfrac{140}{5} \\[5ex] x = 28 \\[3ex] A \:\:base\:\: \angle = 2x + 30 \\[3ex] Base\:\: \angle = 2(28) + 30 \\[3ex] = 56 + 30 \\[3ex] = 86^\circ $
(5.) ACT Points $A$, $B$, and $C$ are vertices of an equilateral triangle.
Points $A$, $B$, and $D$ are collinear points, with $B$ between $A$ and $D$.
What is the measure of $\angle CBD$?

$ A.\:\: 30^\circ \\[3ex] B.\:\: 40^\circ \\[3ex] C.\:\: 60^\circ \\[3ex] D.\:\: 90^\circ \\[3ex] E.\:\: 120^\circ \\[3ex] $

Points $A$, $B$, and $D$ are collinear points means that points $A$, $B$, and $D$ lie on the same straight line
Let us draw a diagram of the question
Number 5

$ \angle ABC = \angle BCA = \angle CAB = 60^\circ ...\angle \:\:in\:\: equilateral\:\: \triangle \\[3ex] $ We can $\angle CBD$ in two ways
Use any method you like

$ \underline{First\:\: method} \\[3ex] \angle CBD + \angle CBA = 180^\circ ... \angle s \:\:in\:\: a\:\: straight\:\: line \\[3ex] \angle CBD + 60 = 180 \\[3ex] \angle CBD = 180 - 60 \\[3ex] \angle CBD = 120^\circ \\[3ex] \underline{Second\:\: method} \\[3ex] \angle CBD = \angle BAC + \angle BCA ... exterior\:\: \angle \:\:of\:\: \triangle ABC \:\:is\:\: the\:\: sum\:\: of\:\: the \:\:two \:\:interior\:\: opposite\:\: \angle s \\[3ex] \angle CBD = 60 + 60 \\[3ex] \angle CBD = 120^\circ $
(6.) ACT Given that $\angle R$ is the included angle between the 2 congruent sides of the isosceles triangle $\triangle RST$, and the measure of $\angle R$ is $50^\circ$, what is the measure of $\angle S$?

$ A.\;\; 20^\circ \\[3ex] B.\;\; 50^\circ \\[3ex] C.\;\; 65^\circ \\[3ex] D.\;\; 80^\circ \\[3ex] E.\;\; 130^\circ \\[3ex] $

Number 6

$ \triangle RST \\[3ex] \angle RST = \angle RTS = p ... base \angle s \;\;of\;\;isos \;\;\triangle \\[3ex] p + p + 50 = 180 ...sum\;\;of\;\; \angle s \;\;in\;\;a\;\;triangle \\[3ex] 2p = 180 - 50 \\[3ex] 2p = 130 \\[3ex] p = \dfrac{130}{2} \\[5ex] p = 65^\circ $
(7.) JAMB Number 7
In the figure above, PQR is a straight line segment.
$|PQ| = |QT|$
Triangle PQT is an isosceles triangle.
$\angle SRQ$ is $75^\circ$ and $\angle QRT$ is $25^\circ$
Calculate the value of $\angle RST$

$ A.\;\; 45^\circ \\[3ex] B.\;\; 55^\circ \\[3ex] C.\;\; 25^\circ \\[3ex] D.\;\; 50^\circ \\[3ex] $

$ \angle QPT = \angle QTP = 25^\circ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle \\[3ex] \angle RQS = 25 + 25 = 50^\circ ...exterior\;\;\angle\;\;of\;\;\triangle PTQ \:\:is\:\: the\:\: sum\:\: of\:\: the \:\:two \:\:interior\:\: opposite\:\: \angle s \\[3ex] \angle RST + 75 + 50 = 180 ... sum\;\;of\;\;\angle s\;\;of\;\;\triangle QSR \\[3ex] \angle RST + 125 = 180 \\[3ex] \angle RST = 180 - 125 \\[3ex] \angle RST = 55^\circ $
(8.) ACT In $\triangle ABC$ shown below, the given side lengths are in meters.
Which of the following expressions gives the area, in square meters, of $\triangle ABC$?
Number 8

$ F.\;\; \dfrac{1}{2}(11)(10) \\[5ex] G.\;\; \sqrt{10^2 + 11^2} \\[3ex] H.\;\; \dfrac{1}{2}(11)(10)(\cos 30^\circ) \\[5ex] J.\;\; \dfrac{1}{2}(11)(10)(\sin 30^\circ) \\[5ex] K.\;\; \sqrt{10^2 + 11^2 - 2(10)(11)(\cos 30^\circ)} \\[3ex] $

$ Area = \dfrac{1}{2} * firstside * secondside * \sin (thirdAngle) \\[5ex] Area = \dfrac{1}{2} * firstside * thirdside * \sin (secondAngle) \\[5ex] Area = \dfrac{1}{2} * secondside * thirdside * \sin (firstAngle) \\[5ex] \therefore Area = \dfrac{1}{2}(11)(10)(\sin 30^\circ) $
(9.) JAMB NUmber 9
Find the value of $\theta$ in the diagram above.

$ A.\;\; 100^\circ \\[3ex] B.\;\; 120^\circ \\[3ex] C.\;\; 30^\circ \\[3ex] D.\;\; 60^\circ \\[3ex] $

$ (\sqrt{3}t)^2 = t^2 + t^2 - 2(t)(t) \cos \theta...Cosine\;\;Law \\[3ex] 3t^2 = 2t^2 - 2t^2 \cos\theta \\[3ex] 3t^2 + 2t^2 \cos\theta = 2t^2 \\[3ex] 2t^2 \cos\theta = 2t^2 - 3t^2 \\[3ex] 2t^2 \cos\theta = -t^2 \\[3ex] \cos\theta = -\dfrac{-t^2}{2t^2} \\[5ex] \cos\theta = -\dfrac{1}{2} \\[5ex] \theta = \cos^{-1} \left(-\dfrac{1}{2}\right) \\[5ex] 1st:2nd:3rd:4th \implies A:S:T:C \\[3ex] cosine\;\;is\;\;negative\;\;in\;\;2nd \\[3ex] Eliminate\;\;Options\;\;C\;\;and\;\;D\;\;because\;\;they\;\;are\;\;in\;\;1st \\[3ex] Option\;\;A\;\;is\;\;not\;\;a\;\;special\;\;angle \\[3ex] Option\;\;B\;\;is\;\;the\;\;answer $
(10.) ACT In $\triangle ABC$, the sum of the measures of $\angle A$ and $\angle B$ is $64^\circ$.
What is the measure of $\angle C$?

$ F.\;\; 26^\circ \\[3ex] G.\;\; 52^\circ \\[3ex] H.\;\; 64^\circ \\[3ex] J.\;\; 116^\circ \\[3ex] K.\;\; 128^\circ \\[3ex] $

$ \angle A + \angle B + \angle C = 180^\circ ...Sum\:\: of\:\: \angle s \:\:of \:\:a\:\: triangle \\[3ex] \angle A + \angle B = 64^\circ \\[3ex] \implies 64 + \angle C = 180 \\[3ex] \angle C = 180 - 64 \\[3ex] \angle C = 116^\circ $
(11.) JAMB A point P moves such that it is equidistant from points Q and R.
Find $|QR|$ when $|PR| = 8\;cm$ and $\angle PRQ = 30^\circ$

$ A.\;\; 4\sqrt{3}\;cm \\[3ex] B.\;\; 8\;cm \\[3ex] C.\;\; 8\sqrt{3}\;cm \\[3ex] D.\;\; 4\;cm \\[3ex] $

Let us draw a diagram of the question.
Number 11

$ |PR| = |QR| = 8\;cm...isosceles\;\;\triangle \\[3ex] \angle PRQ = \angle PQR = 30^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;triangle \\[3ex] \angle QPR + 30 + 30 = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle QPR = 180 - 30 - 30 \\[3ex] \angle QPR = 120^\circ \\[3ex] |QR|^2 = 8^2 + 8^2 - 2(8)(8) \cos 120...Cosine\;\;Law \\[3ex] \cos 120 = -\cos (180 - 120) = -\cos 60 = -\dfrac{1}{2} ... Second\;\;Quadrant\;\;Identity \\[5ex] |QR|^2 = 64 + 64 - 128 * -\dfrac{1}{2} \\[5ex] |QR|^2 = 128 + 64 \\[3ex] |QR|^2 = 192 \\[3ex] |QR| \\[3ex] = \sqrt{192} \\[3ex] = \sqrt{4} * \sqrt{48} \\[3ex] = 2 * \sqrt{16} * \sqrt{3} \\[3ex] = 2 * 4 * \sqrt{3} \\[3ex] = 8\sqrt{3}\;cm $
(12.) ACT In $\triangle ABC,\: \angle A$ measures greater than $22^\circ$ and $\angle B$ measures exactly $58^\circ$.
Which of the following phrases best describes the measure of $\angle C$?
A. Greater than $100^\circ$
B. Less than $100^\circ$
C. Equal to $60^\circ$
D. Equal to $80^\circ$
E. Equal to $100^\circ$


$ \angle A + \angle B + \angle C = 180 ...sum \:\:of\:\: \angle s \:\:in\:\: a\:\: \triangle \\[3ex] \angle A \:\:greater\:\: than\:\: 22 \\[3ex] Assume\:\: \angle A = 23 \\[3ex] \angle B = 58 \\[3ex] 23 + 58 + \angle C = 180 \\[3ex] 81 + \angle C = 180 \\[3ex] \angle C = 180 - 81 = 99 \\[3ex] \implies \angle C \:\:less\:\: than\:\: 100^\circ $
(13.) ACT Radius $\overline{OA}$ of the circle shown below is perpendicular to $\overline{AP}$
The circle intersects $\overline{OP}$ at B.
The length of $\overline{AP}$ is 12 centimeters, and the measure of $\angle APO$ is $20^\circ$
Which of the following values is closest to the length, in centimeters, of $\overline{BP}$?
NUmber 13

(Note: $\sin 20^\circ \approx 0.342$, $\cos 20^\circ \approx 0.940$, and $\tan 20^\circ \approx 0.364$)

$ A.\;\; 2.1 \\[3ex] B.\;\; 4.4 \\[3ex] C.\;\; 6.9 \\[3ex] D.\;\; 7.6 \\[3ex] E.\;\; 8.4 \\[3ex] $

$ SOH:CAH:TOA \\[3ex] \tan 20 = \dfrac{opp}{adj} = \dfrac{\overline{OA}}{12} \\[5ex] \overline{OA} = 12 * \tan 20 \\[3ex] \overline{OA} = 12 * 0.364 \\[3ex] \overline{OA} = 4.368 \\[3ex] \overline{OA} = \overline{OB} = 4.368 ...base\;\;\angle s\;\;of\;\;isosceles \triangle \\[3ex] Considering\;\;\triangle OAP \\[3ex] \overline{OP}^2 = \overline{OA}^2 + \overline{AP}^2 ... Pythagorean\:\: Theorem \\[3ex] \overline{OP}^2 = 4.368^2 + 12^2 \\[3ex] \overline{OP}^2 = 19.079424 + 144 \\[3ex] \overline{OP}^2 = 163.079424 \\[3ex] \overline{OP} = \sqrt{163.079424} \\[3ex] \overline{OP} = 12.77025544 \\[3ex] \overline{OP} = \overline{OB} + \overline{BP} ... based\;\;on\;\;the\;\;diagram \\[3ex] 12.77025544 = 4.368 + \overline{BP} \\[3ex] 4.368 + \overline{BP} = 12.77025544 \\[3ex] \overline{BP} = 12.77025544 - 4.368 \\[3ex] \overline{BP} = 8.40225544 \\[3ex] \overline{BP} \approx 8.4\;cm $
(14.) ACT For $\triangle ABC$ shown below, the length of $\overline{BC}$ is 50 mm
Which of the following equations, when solved, will give the length, in millimeters, of $\overline{AB}$?
(Note: The law of sines states that given $\triangle XYZ$, $\dfrac{\sin \angle X}{YZ} = \dfrac{\sin \angle Y}{XZ} = \dfrac{\sin \angle Z}{XY}$)

Number 14
$ F.\;\; \dfrac{\sin 68^\circ}{50} = \dfrac{\sin 58^\circ}{AB} \\[5ex] G.\;\; \dfrac{\sin 58^\circ}{50} = \dfrac{\sin 68^\circ}{AB} \\[5ex] H.\;\; \dfrac{\sin 58^\circ}{50} = \dfrac{\sin 54^\circ}{AB} \\[5ex] J.\;\; \dfrac{\sin 54^\circ}{50} = \dfrac{\sin 68^\circ}{AB} \\[5ex] K.\;\; \dfrac{\sin 54^\circ}{50} = \dfrac{\sin 58^\circ}{AB} \\[5ex] $

$ \dfrac{\sin C}{?} = \dfrac{\sin 58}{50} \\[5ex] \dfrac{\sin 68}{\overline{AB}} = \dfrac{\sin 58}{50} \\[5ex] Option\;\;G $
(15.) ACT A hill makes an angle of $20^\circ$ with the horizontal, $\overrightarrow{AD}$, as shown below.
A taut guy wire, $\overrightarrow{AB}$, extends from the base of the hill, point A, to point B on a vertical pole.
Point B is 25 ft directly above where the pole is inserted into the ground at point C
Given that the length of $\overline{AC}$ is 60 ft, which of the following expressions represents the length, in feet, of the guy wire?
(Note: For a triangle with sides of length a, b and c that are opposite angles $\angle A$, $\angle B$, and $\angle C$, respectively)
$\dfrac{\sin \angle A}{a} = \dfrac{\sin \angle B}{b} = \dfrac{\sin C}{c}$ and $c^2 = a^2 + b^2 - 2ab \cos \angle C$
NUmber 15

$ F.\;\; \dfrac{25\sin 60^\circ}{\sin 20^\circ} \\[5ex] G.\;\; \dfrac{25\sin 70^\circ}{\sin 20^\circ} \\[5ex] H.\;\; \dfrac{25\sin 110^\circ}{\sin 20^\circ} \\[5ex] J.\;\; \sqrt{60^2 + 25^2 - 2(60)(25) \cos 70^\circ} \\[3ex] K.\;\; \sqrt{60^2 + 25^2 - 2(60)(25) \cos 110^\circ} \\[3ex] $

$ \angle ADC = 180 - 90 ... \angle s \;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle ADC = 90^\circ \\[3ex] \angle ACB = \angle CAD + \angle ADC ... exterior\;\; \angle\;\;of\;\;\triangle ACD \:\:is\:\: the\:\: sum\:\: of\:\: the \:\:two \:\:interior\:\: opposite\:\: \angle s \\[3ex] \angle ACB = 20 + 90 \\[3ex] \angle ACB = 110^\circ \\[3ex] \triangle ABC \\[3ex] a = 25' \\[3ex] b = 60' \\[3ex] \angle ACB = \angle C = 110^\circ \\[3ex] length\;\;of\;\;the\;\;guy\;\;wire = c \\[3ex] c^2 = a^2 + b^2 - 2ab \cos \angle C...Cosine\;\;Law \\[3ex] c^2 = 25^2 + 60^2 - 2(25)(60) \cos 110 \\[3ex] c = \sqrt{25^2 + 60^2 - 2(25)(60) \cos 110} $
(16.)

(17.) CSEC The diagram below, not drawn to scale, shows triangle MNP in which angle MPN = angle PMN = $52^\circ$ and MN = 12.5 cm
Number 17

(i) State the type of triangle shown above.
(ii) Determine the value of angle PNM


(i)
The triangle is an isosceles triangle because $\angle PMN = \angle NMP = 52^\circ$...base angles of isosceles triangle

$ (ii) \\[3ex] 52 + 52 + \angle PNM = 180 ...sum\;\;of\;\;angles\;\;of\;\;\triangle PNM \\[3ex] 104 + \angle PNM = 180 \\[3ex] \angle PNM = 180 - 104 \\[3ex] \angle PNM = 76^\circ \\[3ex] (iii) \\[3ex] Because: \\[3ex] \angle PMN = \angle NMP = 52^\circ \\[3ex] MN = PN = 12.5\;cm...isosceles\;\;\triangle PNM \\[3ex] Area\;\;of\;\;\triangle MNP = \dfrac{1}{2} * MN * PN * \sin \angle PNM \\[5ex] = 0.5 * 12.5 * 12.5 * \sin 76 \\[5ex] = 78.125 * 0.9702957263 \\[3ex] = 75.80435362 \\[3ex] \approx 75.8\;cm^2 $
(18.) ACT In $\triangle ABC$, the measure of $\angle A$ is $43^\circ$ and the measure of $\angle C$ is $32^\circ$
Which of the following inequalities involving the lengths of the sides of $\triangle ABC$ is true?

$ A.\;\; AB \gt AC \\[3ex] B.\;\; AB \gt BC \\[3ex] C.\;\; AC \gt BC \\[3ex] D.\;\; AC \gt AB + BC \\[3ex] E.\;\; BC \gt AC \\[3ex] $

First: we need to draw the triangle.
Number 18

$ \angle A + \angle B + \angle C = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;\triangle ABC \\[3ex] 43 + \angle B + 32 = 180 \\[3ex] \angle B + 75 = 180 \\[3ex] \angle B = 180 - 75 \\[3ex] \angle B = 105 \\[3ex] $ Second: We apply one of the Triangle Theorems
Triangle Theorem Number (10.)
The measure of an angle is proportional to the length of the side; side length is proportional to angle measure.
Therefore:

$ 105^\circ \gt 43^\circ \gt 32^\circ \\[3ex] B^\circ \gt A^\circ \gt C^\circ \\[3ex] \implies \\[3ex] b \gt a \gt c \\[3ex] AC \gt BC \gt AB \\[3ex] AC \gt BC ...Option\;\;C $
(19.) ACT In the figure below, $\overleftrightarrow{AD}$ intersects $\overleftrightarrow{BG}$ at $C$ and is perpendicular to $\overleftrightarrow{DE}$
Line $\overleftrightarrow{DE}$ intersects $\overleftrightarrow{BG}$ at $F$
Given that the measure of $\angle EFG$ is $25^\circ$, what is the measure of $\angle BCD$?
Number 19

$ F.\;\; 65^\circ \\[3ex] G.\;\; 115^\circ \\[3ex] H.\;\; 120^\circ \\[3ex] J.\;\; 130^\circ \\[3ex] K.\;\; 155^\circ \\[3ex] $

$ \angle CFD = 25^\circ ...Vertical\;\;\angle s\;\;are\;\;equal \\[3ex] \angle BCD = \angle CFD + \angle CDF ...exterior\;\;angle\;\;of\;\;\triangle CDF\;\;is\;\;the\;\;sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;angles \\[3ex] \angle BCD = 25 + 90 \\[3ex] \angle BCD = 115^\circ $
(20.) ACT In the figure below, $C$ is on $\overline{BD}$, $\angle BAC$ measures $42^\circ$, and $\angle ABC$ measures $108^\circ$
What is the measure of $\angle ACD$?
Number 20

$ A.\;\; 108^\circ \\[3ex] B.\;\; 120^\circ \\[3ex] C.\;\; 132^\circ \\[3ex] D.\;\; 138^\circ \\[3ex] E.\;\; 150^\circ \\[3ex] $

$ \angle ACD = \angle BAC + \angle ABC \\[3ex] ...exterior\;\;\angle \;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \implies \\[3ex] \angle ACD = 42^\circ + 108^\circ \\[3ex] \angle ACD = 150^\circ $




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(21.) WASSCE A ladder $11\;m$ long leans against a vertical wall at an angle of $75^\circ$ to the ground.
The ladder is pushed $0.2\;m$ up the wall.

(a.) Illustrate the information in a diagram.

(b.) Find, correct to the nearest whole number, the:
(i.) new angle which the ladder makes with the ground
(ii.) distance the foot of the ladder has moved from its original position.


(a.)
Number 21

$ (b.) \\[3ex] (i.) \\[3ex] SOH:CAH:TOA \\[3ex] \underline{\triangle ABC} \\[3ex] \sin 75 = \dfrac{BC}{11} \\[5ex] BC = 11 \sin 75 \\[3ex] BC = 11(0.9659258263) \\[3ex] BC = 10.62518409 \\[3ex] \underline{Both\;\;\triangle s} \\[3ex] BD = BC + CD \\[3ex] BD = 10.62518409 + 0.2 \\[3ex] BD = 10.82518409 \\[3ex] \underline{\triangle EBD} \\[3ex] \sin \theta = \dfrac{BD}{11} \\[5ex] \sin \theta = \dfrac{10.82518409}{11} \\[5ex] \sin \theta = 0.9841076445 \\[3ex] \theta = \sin^{-1}(0.9841076445) \\[3ex] \theta = 79.77157858 \\[3ex] \theta \approx 80^\circ \\[3ex] (ii.) \\[3ex] \underline{\triangle ABC} \\[3ex] \cos 75 = \dfrac{AB}{11} \\[5ex] AB = 11\cos 75 \\[3ex] AB = 11(0.2588190451) \\[3ex] AB = 2.847009496 \\[3ex] \underline{\triangle EBD} \\[3ex] \cos \theta = \dfrac{EB}{11} \\[5ex] EB = 11\cos \theta \\[3ex] EB = 11\cos 79.77157858 \\[3ex] EB = 11(0.1775729261) \\[3ex] EB = 1.953302188 \\[3ex] \underline{Both\;\; \triangle s} \\[3ex] AE + EB = AB \\[3ex] AE = AB - EB \\[3ex] AE = 2.847009496 - 1.953302188 \\[3ex] AE = 0.8937073084 \\[3ex] AE \approx 1\;m $
(22.) JAMB
Number 22

The traingle PQR above is
A. an isosceles triangle
B. an obtuse-angled triangle
C. a scalene triangle
D. an equilateral triangle


$ \angle PRQ + 128 = 180...\angle s \;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle PRQ = 180 - 128 \\[3ex] \angle PRQ = 52^\circ \\[3ex] \angle QPR + 76 + 52 = 180...sum\;\;of\;\;\angle s\;\;of\;\;\triangle PQR \\[3ex] \angle QPR + 128 = 180 \\[3ex] \angle QPR = 180 - 128 \\[3ex] \angle QPR = 52^\circ \\[3ex] Because\;\; \angle QPR = \angle QRP = 52^\circ: \\[3ex] \triangle PQR \;\;is\;\;an\;\;isosceles\;\; \triangle $
(23.) ACT A forest fire is contained within a triangular region, which is shown below.
The supervising firefighter plans to fight the fire by positioning a firefighter about every 4 meters along the perimeter of the triangle.
Among the following, which expression best estimates the planned number of firefighters along the perimeter?
(Note: The law of sines states that in every triangle, the 3 ratios of length of a side to the sine of the angle opposite that side are equal.)
Number 23

$ A.\;\; \dfrac{130 + \left(\dfrac{130\sin 42^\circ}{\sin 91^\circ}\right) + \left(\dfrac{130\sin 47^\circ}{\sin 91^\circ}\right)}{4} \\[7ex] B.\;\; \dfrac{130 + \left(\dfrac{130\sin 91^\circ}{\sin 42^\circ}\right) + \left(\dfrac{130\sin 91^\circ}{\sin 47^\circ}\right)}{4} \\[7ex] C.\;\; 130 + \dfrac{130 \sin 42^\circ}{\sin 91^\circ} + \dfrac{130\sin 47^\circ}{\sin 91^\circ} \\[5ex] D.\;\; \dfrac{\dfrac{1}{2}\left(\dfrac{130\sin 47^\circ}{\sin 91^\circ}\right)}{4} \\[7ex] E.\;\; \dfrac{\dfrac{1}{2}(130)}{4} \\[5ex] $

$ Because\;\;of:\;\;positioning\;\;a\;\;firefighter\;\;about\;\;every\;\;4\;\;meters\;\;along\;\;the\;\;perimeter\;\;of\;\;the\;\; triangle \\[3ex] Planned\;\;number\;\;of\;\;firefighters = \dfrac{Perimeter}{4} \\[5ex] $
Number 23

$ \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \\[5ex] \dfrac{a}{\sin 47} = \dfrac{130}{91} ...Sine\;\;Law \\[5ex] a = \dfrac{130 \sin 47}{\sin 91} \\[5ex] \angle ACB + 47 + 91 = 180...Sum\;\;\angle s\;\;of\;\;\triangle ABC \\[3ex] \angle ACB = 180 - 47 - 91 \\[3ex] \angle ACB = 42 \\[3ex] \dfrac{c}{\sin 42} = \dfrac{130}{91} ...Sine\;\;Law \\[5ex] c = \dfrac{130 \sin 42}{\sin 91} \\[5ex] Perimeter \\[3ex] = a + b + c \\[3ex] = \dfrac{130 \sin 47}{\sin 91} + 130 + \dfrac{130 \sin 42}{\sin 91} \\[5ex] \therefore \;\;Planned\;\;number\;\;of\;\;firefighters \\[3ex] = \dfrac{\dfrac{130 \sin 47}{\sin 91} + 130 + \dfrac{130 \sin 42}{\sin 91}}{4} \\[7ex] = \dfrac{130 + \left(\dfrac{130\sin 42^\circ}{\sin 91^\circ}\right) + \left(\dfrac{130\sin 47^\circ}{\sin 91^\circ}\right)}{4} $
(24.)

(25.) NZQA In this spider web, ABC is a straight line.
Lines AD and BE are parallel to each other and DB = AB
Angle DBE = 64°

Number 25

Calculate the size, p, of angle EBC.
Justify your answer.


construction: Draw the line from point D to point E to use the angle properties between parallel lines
Number 25

$ \angle ADB = 64^\circ ...alternate \;\;\angle s\;\;are\;\;equal \\[3ex] \angle DAB = \angle ADB = 64^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle ADB \\[3ex] \angle DBC = \angle ADB + \angle DAB ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle DBC = 64 + p ...diagram \\[3ex] \implies \\[3ex] 64 + p = 64 + 64 \\[3ex] \therefore p = 64^\circ $
(26.) JAMB In triangle $PQR$, $q = 8\;cm$, $r = 6\;cm$ and $\cos P = \dfrac{1}{12}$
Calculate the value of $p$

$ A.\;\; 10\;cm \\[3ex] B.\;\; \sqrt{108}\;cm \\[3ex] C.\;\; 9\;cm \\[3ex] D.\;\; \sqrt{92}\; cm \\[3ex] $

Nothing in the question mentions that the triangle is a right triangle (nothing in the question tells us that one of the angles in the triangle is a right angle)
Hence, we shall draw a triangle that is not a right triangle
Let us draw the diagram for the question
Number 26

$ \underline{\triangle PQR} \\[3ex] q = 8\;cm \\[3ex] r = 6\;cm \\[3ex] \cos P = \dfrac{1}{12} \\[5ex] p^2 = q^2 + r^2 - 2qr \cos P ...Cosine\;\;Law \\[3ex] p^2 = 8^2 + 6^2 - 2(8)(6) * \dfrac{1}{12} \\[5ex] p^2 = 64 + 36 - 96 * \dfrac{1}{12} \\[5ex] p^2 = 100 - 8 \\[3ex] p^2 = 92 \\[3ex] p = \sqrt{92} $
(27.) WASSCE:FM The lengths of the sides of a triangle, in centimeters, are $(3 + \sqrt{2})$, $2\sqrt{3}$ and $3\sqrt{2}$
Find the value of the largest angle of the triangle.


$ 2\sqrt{3} = 3.464101615 \\[3ex] 3\sqrt{2} = 4.242640687 \\[3ex] 3 + \sqrt{2} = 4.414213562 \\[3ex] 3 + \sqrt{2} \gt 3\sqrt{2} \gt 2\sqrt{3} \\[3ex] $ Based on the Triangle Theorem Number (10.)
The measure of an angle is proportional to the length of the side; side length is proportional to angle measure.
This implies that the value of the largest angle of the triangle is the angle facing $3 + \sqrt{2}$
Let that angle be $\theta$
Number 27
$ (3 + \sqrt{2})^2 = (3\sqrt{2})^2 + (2\sqrt{3})^2 - 2(3\sqrt{2})(2\sqrt{3}) \cos \theta ...Cosine\;\;Law \\[3ex] (3 + \sqrt{2})^2 + 2(3\sqrt{2})(2\sqrt{3}) \cos \theta = (3\sqrt{2})^2 + (2\sqrt{3})^2 \\[3ex] 2(3\sqrt{2})(2\sqrt{3}) \cos \theta = (3\sqrt{2})^2 + (2\sqrt{3})^2 - (3 + \sqrt{2})^2 \\[3ex] 2 * 3 * \sqrt{2} * 2 * \sqrt{3} * \cos \theta = (3^2 * \sqrt{2}^2) + (2^2 * \sqrt{3}^2) - [(3 + \sqrt{2})(3 + \sqrt{2})] \\[3ex] 12\sqrt{6}\cos\theta = 9(2) + 4(3) - (9 + 3\sqrt{2} + 3\sqrt{2} + \sqrt{2}^2) \\[3ex] 12\sqrt{6}\cos\theta = 18 + 12 - (9 + 6\sqrt{2} + 2) \\[3ex] 12\sqrt{6}\cos\theta = 30 - (11 + 6\sqrt{2}) \\[3ex] 12\sqrt{6}\cos\theta = 30 - 11 - 6\sqrt{2} \\[3ex] 12\sqrt{6}\cos\theta = 19 - 6\sqrt{2} \\[3ex] \cos\theta = \dfrac{19 - 6\sqrt{2}}{12\sqrt{6}} \\[5ex] \cos\theta = \dfrac{19 - 8.485281374}{29.39387691} \\[5ex] \cos\theta = \dfrac{10.51471863}{29.39387691} \\[5ex] \cos \theta = 0.357717992 \\[3ex] \theta = \cos^{-1} (0.3557717992) \\[3ex] \theta = 69.03988392^\circ $
(28.) ACT A triangle has sides of length 2.5 feet and 4 feet.
Which of the following CANNOT be the length of the third side, in feet?

$ F.\;\; 1 \\[3ex] G.\;\; 2 \\[3ex] H.\;\; 3 \\[3ex] J.\;\; 4 \\[3ex] K.\;\; 5 \\[3ex] $

$ \underline{Triangle\;\;Inequality\;\;Theorem} \\[3ex] Option\;\:F \\[3ex] 1 + 2.5 \lt 4 \\[3ex] 3.5 \lt 4 ...No,\;\;it\;\;cannot...Correct\;\;Option \\[3ex] Option\;\; G \\[3ex] 2 + 2.5 \gt 4 \\[3ex] 4.5 \gt 4...Yes,\;\;it\;\;can \\[3ex] Option\;\; H \\[3ex] 3 + 2.5 \gt 4 \\[3ex] 5.5 \gt 4...Yes,\;\;it\;\;can \\[3ex] Option\;\; J \\[3ex] 4 + 2.5 \gt 4 \\[3ex] 6.5 \gt 4...Yes,\;\;it\;\;can \\[3ex] Option\;\; H \\[3ex] 5 + 2.5 \gt 4 \\[3ex] 7.5 \gt 4...Yes,\;\;it\;\;can $
(29.) CSEC In the diagram below, not drawn to scale, $RQ = 9\;m$, $RS = 12\;m$, $ST = 13\;$, $\angle QRS = 60^\circ$ and $\angle SQT = 40^\circ$

Number 29

Calculate, correct to 1 decimal place,
(i.) the length $QS$
(ii.) the measure of $\angle QTS$
(iii.) the area of triangle $QRS$
(iv.) the perpendicular distance from $Q$ to $RS$


$ (i.) \\[3ex] \underline{\triangle QRS} \\[3ex] |QS|^2 = |QR|^2 + |RS|^2 - 2(|QR|)(|RS|) * \cos \angle QRS ...Cosine\;\;Law \\[3ex] |QS|^2 = 9^2 + 12^2 - 2(9)(12) * \cos 60^\circ \\[3ex] |QS|^2 = 81 + 144 - 216 * \dfrac{1}{2} \\[5ex] |QS|^2 = 225 - 108 \\[3ex] |QS|^2 = 117 \\[3ex] |QS| = \sqrt{117} \\[3ex] |QS| = 10.81665383 \\[3ex] |QS| \approx 10.8\;m \\[3ex] (ii.) \\[3ex] \underline{\triangle QST} \\[3ex] \dfrac{\sin \angle QTS}{|QS|} = \dfrac{\sin SQT}{|ST|}...Sine\;\;Law \\[5ex] \dfrac{\sin \angle QTS}{10.81665383} = \dfrac{\sin 40}{13} \\[5ex] \sin \angle QTS = \dfrac{\sin 40 * 10.81665383}{13} \\[5ex] \sin \angle QTS = \dfrac{0.6427876097 * 10.81665383}{13} \\[5ex] \sin \angle QTS = \dfrac{6.95281106}{13} \\[5ex] \sin \angle QTS = 0.53483162 \\[3ex] \angle QTS = \sin^{-1}(0.53483162) \\[3ex] \angle QTS = 32.33249302 \\[3ex] \angle QTS \approx 32.3^\circ \\[3ex] (iii.) \\[3ex] Area\;\;of\;\;\triangle QRS \\[3ex] = \dfrac{1}{2} * |QR| * |RS| * \sin \angle QRS \\[5ex] = \dfrac{1}{2} * 9 * 12 * \sin 60 \\[5ex] = 9 * 6 * 0.8660254038 \\[3ex] = 46.7653718 \\[3ex] \approx 46.8\;m^2 \\[3ex] (iv.) \\[3ex] Perpendicular\;\;distance\;\;from\;\;Q\;\;to\;\;RS = |QP| \\[3ex] $
Number 29

$ \underline{\triangle RQP} \\[3ex] SOH:CAH:TOA \\[3ex] \sin 60 = \dfrac{opp}{hyp} = \dfrac{|QP|}{9} \\[5ex] 9 \sin 60 = |QP| \\[3ex] |QP| = 9\sin 60 \\[3ex] |QP| = 9(0.8660254038) \\[3ex] |QP| = 7.794228634 \\[3ex] |QP| \approx 7.8\;m $
(30.) curriculum.gov.mt The angles of a triangle are $(x + 10)^\circ$, $(2x + 14)^\circ$ and $(x - 8)^\circ$

Number 30

(a) Write an expression, in terms of x, for the sum of the angles.
Give your answer in its simplest form.

(b) The sum of the angles in a triangle is 180°
(i) Use your answer in part (a) to write an equation in x and solve it.
(ii) Use your answer in part (b)(i) to work out the size of angle $A\hat{B}C$


$ (a) \\[3ex] Sum\;\;of\;\;the\;\;\angle s \\[3ex] = (x + 10) + (2x + 14) + (x - 8) \\[3ex] = x + 10 + 2x + 14 + x - 8 \\[3ex] = 4x + 16 \\[3ex] = 4(x + 4) \\[3ex] (b) \\[3ex] (i) \\[3ex] 4(x + 4) = 180 \\[3ex] x + 4 = \dfrac{180}{4} \\[5ex] x + 4 = 45 \\[3ex] x = 45 - 4 \\[3ex] x = 41^\circ \\[3ex] (ii) \\[3ex] A\hat{B}C \\[3ex] = 2x + 14 \\[3ex] = 2(41) + 14 \\[3ex] = 82 + 14 \\[3ex] = 96^\circ $
(31.)

(32.) ACT The perimeter of a certain scalene triangle is 100 inches.
The side lengths of the triangle are represented by $5x$, $3x + 30$, and $2x + 10$, respectively.
What is the length, in inches, of the longest side of the triangle?

$ A.\;\; 6 \\[3ex] B.\;\; 22 \\[3ex] C.\;\; 30 \\[3ex] D.\;\; 48 \\[3ex] E.\;\; 72 \\[3ex] $

$ 5x + (3x + 30) + (2x + 10) = 100 \\[3ex] 5x + 3x + 30 + 2x + 10 = 100 \\[3ex] 10x + 40 = 100 \\[3ex] 10x = 100 - 40 \\[3ex] 10x = 60 \\[3ex] x = \dfrac{60}{10} \\[5ex] x = 6 \\[3ex] 1st\;\;side = 5x = 5(6) = 30\;inches \\[3ex] 2nd\;\;side = 3x + 30 = 3(6) + 30 = 18 + 30 = 48\;inches \\[3ex] 3rd\;\;side = 2x + 10 = 2(6) + 10 = 12 + 10 = 22\;inches \\[3ex] Longest\;\;side = 48\;inches $
(33.) ACT In the figure below, the distances between 2 pairs of cities are shown, as well as the angle formed at Ewing, which has a measure of 127°
Which of the following values is closest to the distance, in miles, from Deerborn to Fergus?
Number 33

(Note: $\cos 127^\circ = -0.6$; $\sin 127^\circ = 0.8$)

$ A.\;\; 100 \\[3ex] B.\;\; 140 \\[3ex] C.\;\; 160 \\[3ex] D.\;\; 180 \\[3ex] E.\;\; 200 \\[3ex] $

Let the distance from Deerborn to Fergus = e

$ e^2 = 80^2 + 120^2 - 2(80)(120) * \cos 127^\circ ...Cosine\;\;Law \\[3ex] e^2 = 6400 + 14400 - 19200 * -0.6 \\[3ex] e^2 = 20800 + 11520 \\[3ex] e^2 = 32320 \\[3ex] e = \sqrt{32320} \\[3ex] e = 179.7776404 \\[3ex] e \approx 180\;miles $
(34.)

(35.)

(36.) ACT In the parallelogram below, what is the measure of $\angle DAC$ ?

Number 36

$ A.\;\; 20^\circ \\[3ex] B.\;\; 30^\circ \\[3ex] C.\;\; 40^\circ \\[3ex] D.\;\; 50^\circ \\[3ex] E.\;\; 70^\circ \\[3ex] $

$ |DC|\;\; || \;\;|AB| \\[3ex] ... Opposite\;\;sides\;\;of\;\;a\;\;parallelogram\;\;are\;\;|| \\[3ex] \angle DAC = 50^\circ ...alternate\;\;\angle s\;\;are\;\;equal $
(37.)

(38.) curriculum.gov.mt Work out the size of the angle marked a

Number 38


Number 38

$ \underline{\triangle ABC} \\[3ex] \angle ABC = \angle ACB = 55^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle ABC \\[3ex] \angle ABC + \angle ACB + \angle BAC = 180^\circ...sum\;\;of\;\;\angle s\;\;of\;\;\triangle ABC \\[3ex] \implies \\[3ex] 55 + 55 + a = 180 \\[3ex] 110 + a = 180 \\[3ex] a = 180 - 110 \\[3ex] a = 70^\circ $
(39.) ACT In the figure below, lines j and k are parallel, lines m and n are parallel, and the measures of 2 angles are shown.
What is the measure of $\angle x$?

Number 39

$ A.\;\; 45^\circ \\[3ex] B.\;\; 55^\circ \\[3ex] C.\;\; 65^\circ \\[3ex] D.\;\; 75^\circ \\[3ex] E.\;\; 80^\circ \\[3ex] $

Number 39

$ \angle ABC = 45^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle BAC = x ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle BCD = \angle ABC + \angle BAC ...exterior\;\;\angle\;\;of\;\;\triangle ABC = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \implies \\[3ex] 100 = 45 + x \\[3ex] 45 + x = 100 \\[3ex] x = 100 - 45 \\[3ex] x = 55^\circ $
(40.)





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(41.) curriculum.gov.mt The height of the isosceles triangle is 30 cm and the base is 20 cm.
What is the value of tan P?

Number 41


Number 41

$ BD = DC = 10\;cm ...height\;\;drawn\;\;from\;\;the\;\;apex\;\;of\;\;an\;\;isosceles\;\;\triangle\;\;bisects\;\;the\;\;base \\[3ex] \tan P = \dfrac{30}{10} ...SOHCAHTOA \\[5ex] \tan P = 3 $
(42.) ACT In $\triangle ABC$ shown below, the measure of $\angle A$ is 58°, and $\overline{AB} = \overline{AC}$
What is the measure of $\angle C$?

Number 42

$ A.\;\; 32^\circ \\[3ex] B.\;\; 42^\circ \\[3ex] C.\;\; 58^\circ \\[3ex] D.\;\; 61^\circ \\[3ex] E.\;\; 62^\circ \\[3ex] $

$ \overline{AB} \cong \overline{AC} \implies \triangle CBA \;\;is\;\;an\;\;isosceles\;\;\triangle \\[3ex] \angle C = \angle B = p...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle CBA \\[3ex] \angle C + \angle B + \angle A = 180^\circ ...sum\;\;of\;\;\angle s\;\;\triangle CBA \\[3ex] p + p + 58 = 180 \\[3ex] 2p = 180 - 58 \\[3ex] 2p = 122 \\[3ex] p = \dfrac{122}{2} \\[5ex] p = 61 \\[3ex] \therefore \angle C = 61^\circ $
(43.) NZQA Straight lines DEF and TMB are parallel to each other.
Lines EM and EB are of equal length.
Angle FEB = 58°

Number 43

Calculate the size, w, of angle MEB.
Justify your answer with clear geometrical reasoning.


$ \angle MBE = 58^\circ...alternate \angle s\;\;are\;\;equal \\[3ex] \angle EMB = \angle MBE = 58^\circ ... base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle MEB \\[3ex] \angle EMB + \angle MBE + \angle MEB = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;\triangle MEB \\[3ex] 58 + 58 + w = 180 \\[3ex] 116 + w = 180 \\[3ex] w = 180 - 116 \\[3ex] w = 64^\circ $
(44.) ACT In $\triangle ABC$ shown below, the given side lengths are in meters.
What is the area, in square meters, of $\triangle ABC$?

Number 44

$ F.\;\; 10 \\[3ex] G.\;\; 12 \\[3ex] H.\;\; 12\sqrt{2} \\[3ex] J.\;\; 12\sqrt{3} \\[3ex] K.\;\; 24 \\[3ex] $

$ \sin 60^\circ = \dfrac{\sqrt{3}}{2} ...Special\;\;\angle \\[5ex] Area \\[3ex] = \dfrac{1}{2} * 6 * 8 * \sin 60^\circ \\[3ex] = 3 * 8 * \dfrac{\sqrt{3}}{2} \\[5ex] = 3 * 4 * \sqrt{3} \\[3ex] = 12\sqrt{3} $
(45.)

(46.)

(47.) ACT The vertices of $\triangle PQR$ are given in the standard (x, y) coordinate plane below.
What is the area, in square coordinate units, of $\triangle PQR$?

Number 47

$ F.\;\; 6 \\[3ex] G.\;\; 8 \\[3ex] H.\;\; 12 \\[3ex] J.\;\; 24 \\[3ex] K.\;\; 28 \\[3ex] $

$ P(5, 11) \\[3ex] Q(8, 7) \\[3ex] R(5, 3) \\[3ex] \underline{Distance\;\;Formula} \\[3ex] |PR| \\[3ex] = \sqrt{(5 - 5)^2 + (11 - 3)^2} \\[3ex] = \sqrt{0^2 + 8^2} \\[3ex] = \sqrt{0 + 64} \\[3ex] = \sqrt{64} \\[3ex] = 8\;units \\[3ex] |PQ| \\[3ex] = \sqrt{(5 - 8)^2 + (11 - 7)^2} \\[3ex] = \sqrt{(-3)^2 + 4^2} \\[3ex] = \sqrt{9 + 16} \\[3ex] = \sqrt{25} \\[3ex] = 5\;units \\[3ex] |QR| \\[3ex] = \sqrt{(8 - 5)^2 + (7 - 3)^2} \\[3ex] = \sqrt{3^2 + 4^2} \\[3ex] = \sqrt{9 + 16} \\[3ex] = \sqrt{25} \\[3ex] = 5\;units \\[3ex] semi-perimeter \\[3ex] = \dfrac{8 + 5 + 5}{2} \\[5ex] = \dfrac{18}{2} \\[5ex] = 9\;units \\[3ex] \underline{Heron's\;\;Formula} \\[3ex] Area \\[3ex] = \sqrt{9(9 - 8)(9 - 5)(9 - 5)} \\[3ex] = \sqrt{9(1)(4)(4)} \\[3ex] = 3(1)(2)(2) \\[3ex] = 12\;square\;units $
(48.)

(49.)

(50.)

(51.) ACT In the figure shown below, E and G lie on $\overline{AC}$, D and F lie on $\overline{AB}$, $\overline{DE}$ and $\overline{FG}$ are parallel to $\overline{BC}$, and the given lengths are in feet.
What is the length of $\overline{AC}$, in feet?

Number 51

$ A.\;\; 13 \\[3ex] B.\;\; 26 \\[3ex] C.\;\; 29 \\[3ex] D.\;\; 42 \\[3ex] E.\;\; 48 \\[3ex] $

Number 51

$ Consider\;\;parallel\;\;lines:\;\; \overline{DE} \;\;and\;\; \overline{BC} \\[3ex] \dfrac{\overline{AD}}{\overline{DB}} = \dfrac{\overline{AE}}{\overline{EC}} \\[5ex] \dfrac{8}{7 + 6} = \dfrac{16}{\overline{EC}} \\[5ex] \dfrac{8}{13} = \dfrac{16}{\overline{EC}} \\[5ex] 8 * \overline{EC} = 13(16) \\[3ex] \overline{EC} = \dfrac{13(16)}{8} \\[5ex] \overline{EC} = 13(2) \\[3ex] \overline{EC} = 26 \\[3ex] \overline{AC} = \overline{AE} + \overline{EC}...figure\;\;shown \\[3ex] \overline{AC} = 16 + 26 \\[3ex] \overline{AC} = 42\;units $
(52.)

(53.)

(54.) ACT In the figure below, C is on $\overline{BD}$, $\angle BAC$ measures 40°, and $\angle ABC$ measures 110°
What is the measure of $\angle ACD$?

Number 54

$ A.\;\; 110^\circ \\[3ex] B.\;\; 120^\circ \\[3ex] C.\;\; 130^\circ \\[3ex] D.\;\; 140^\circ \\[3ex] E.\;\; 150^\circ \\[3ex] $

$ \angle ACD = \angle BAC + \angle ABC \\[3ex] ...exterior\;\;\angle \;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \implies \\[3ex] \angle ACD = 40^\circ + 110^\circ \\[3ex] \angle ACD = 150^\circ $
(55.)

(56.)

(57.)

(58.)



VOYAGERS TO NEW ZEALAND

NZQA
This year (2019) marks 250 years since the first meeting between Māori and the crew of Captain Cook's ship, the HMS Endeavour.
Pacific voyagers settled in New Zealand many years before Captain Cook arrived.
They travelled to New Zealand in double-hulled voyaging canoes, waka hourua, like the one pictured below.

Numbers 59-64

Source: www.maoritelevision.com/news/sport/emerging-navigators-learn-ancient-art-way-finding
The information is related to Questions 59 - 64


(59.) NZQA The canoes had sails.
One of the sails is shown in the diagram below.
BG, CF, and DE are all parallel to each other.
Angle EAD = 44° and angle ADE = 78°

Number 59

Calculate the size, x, of angle AGB.
Justify your answer with clear geometrical reasoning.


$ \angle GAB = \angle EAD = 44^\circ ... diagram \\[3ex] GB || ED \\[3ex] \angle ABG = \angle ADE = 78^\circ ...corresponding\;\;\angle s\;\;are\;\;equal \\[3ex] \angle GAB + \angle ABG + \angle AGB = 180^\circ ... sum\;\;of\;\;\angle s\;\;of\;\;\triangle AGB \\[3ex] 44 + 78 + x = 180 \\[3ex] 122 + x = 180 \\[3ex] x = 180 - 122 \\[3ex] x = 58^\circ $
(60.) NZQA The same sail is shown again below.
The edge of the canoe ABCD is horizontal.
The mast AKH is vertical.
AK = 2.8 metres. KH = 1.2 metres. Angle AKE = 90°.

Number 60

Calculate the area of the sail AHE (shaded grey in the diagram).
Show your working clearly.


$ \angle KAE + \angle EAD = 90^\circ \\[3ex] ... AKH\;(vertical) \perp ABCD\;(horizontal) \\[3ex] \angle KAE + 44 = 90 \\[3ex] \angle KAE = 90 - 44 \\[3ex] \angle KAE = 46^\circ \\[3ex] \underline{\triangle KAE} \\[3ex] \cos KAE = \dfrac{AK}{AE} ...SOHCAHTOA \\[5ex] \cos 46 = \dfrac{2.8}{AE} \\[5ex] AE * \cos 46 = 2.8 \\[3ex] AE = \dfrac{2.8}{\cos 46} \\[5ex] AE = \dfrac{2.8}{0.6946583705} \\[5ex] AE = 4.030758311\;m \\[3ex] \underline{\triangle AHE} \\[3ex] Area = \dfrac{1}{2} * AH * AE * \sin HAE \\[5ex] AH = AK + KH \\[3ex] AH = 2.8 + 1.2 \\[3ex] AH = 4\;m \\[3ex] \angle HAE = \angle KAE = 46^\circ ... diagram \\[3ex] \implies \\[3ex] Area \\[3ex] = \dfrac{1}{2} * 4 * 4.030758311 * \sin 46 \\[5ex] = 2(4.030758311)(0.7193398003) \\[3ex] = 5.798969757\;m^2 $




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(61.) NZQA Tha canoe includes a pair of sails, which are both isosceles triangles, labelled P and Q below.
The angle z is exactly the same size in triangles P and Q.

Number 61

For the canoes to sail as fast as possible, the total area of both sails needs to be greater than 14 $m^2$.
Using the measurements shown in the diagram above, show whether or not the sails satisfy the size requirement.
Justify your answer with clear geometrical reasoning and working.


Because:
The angle z is exactly the same size in triangles P and Q
and:
Triangles P and Q are isosceles
This implies that:
Triangles P and Q have the same angles (same three angles)
Because triangles P and Q have the same angles, triangles P and Q are similar triangles.

$ \underline{\triangle P} \\[3ex] Other\;\;side\;\;length = 6\;m ... isosceles\;\;\triangle \\[3ex] 3.6^2 = 6^2 + 6^2 - 2(6)(6) * \cos z ...Cosine\;\;Law \\[3ex] 2(6)(6) * \cos z = 6^2 + 6^2 - 3.6^2 \\[3ex] 72 * \cos z = 36 + 36 - 12.96 \\[3ex] 72 * \cos z = 59.04 \\[3ex] \cos z = \dfrac{59.04}{72} \\[5ex] \cos z = 0.82 \\[3ex] z = \cos^{-1}(0.82) \\[3ex] z = 34.91520625^\circ \\[3ex] Area \\[3ex] = \dfrac{1}{2} * 6 * 6 * \sin z \\[5ex] = 3(6)(\sin 34.91520625) \\[3ex] = 18(0.5723635209) \\[3ex] = 10.30254338\;m^2 \\[3ex] Let\;\;unknown\;\;equal\;\;side\;\;of\;\;\triangle Q = q \\[3ex] \underline{\sim \triangle s\; P\;\;and\;\;Q} \\[3ex] \dfrac{q}{6} = \dfrac{1.44}{3.6} ...\triangle Q \sim \triangle P \\[5ex] q = \dfrac{6(1.44)}{3.6} \\[5ex] q = \dfrac{8.64}{3.6} \\[5ex] q = 2.4\;m \\[3ex] \underline{\triangle Q} \\[3ex] Area \\[3ex] = \dfrac{1}{2} * 2.4 * 2.4 * \sin z \\[5ex] = 1.2(2.4)(\sin 34.91520625) \\[3ex] = 2.88(0.5723635209) \\[3ex] = 1.64840694\;m^2 \\[3ex] \underline{\triangle s\;P\;\;and\;\;Q} \\[3ex] Total\;\;Area \\[3ex] = 10.30254338 + 1.64840694 \\[3ex] = 11.95095032\;m^2 \\[3ex] $ The total area of both sails is less than 14 $m^2$.
Hence, the sails do not satisfy the size requirement.
(62.) NZQA Captain Cook sailed in a "tall ship" like the one shown in the picture below.

Number 62a

Source: www.telegraph.co.uk/news/2018/09/19/captain-cooks-missing-hms-endeavour-discovered-us-coast/

Part of the sail rigging is shown in the diagram below.

Number 62b

Show that the height of the sail, h, from M to N, is 6.32 metres.
Show your working clearly.


$ \underline{\triangle PNM} \\[3ex] |PN|^2 = |MN|^2 + |PM|^2 ...Pythagorean\;\;theorem \\[3ex] 8^2 = h^2 + 4.9^2 \\[3ex] h^2 + 4.9^2 = 8^2 \\[3ex] h^2 = 8^2 - 4.9^2 \\[3ex] h^2 = 64 - 24.01 \\[3ex] h^2 = 39.99 \\[3ex] h = \sqrt{39.99} \\[3ex] h = 6.323764702 \\[3ex] \approx 6.32\;m $
(63.) NZQA In light winds, the same section of sail expands, as shown below.

Number 63

Calculate the length of the cross-beam, g, from P to Q.
Show your working clearly.


$ \underline{\triangle QNM} \\[3ex] |QN|^2 = |MN|^2 + |QM|^2 ...Pythagorean\;\;theorem \\[3ex] 11^2 = h^2 + (g + 4.9)^2 \\[3ex] h^2 + (g + 4.9)^2 = 11^2 \\[3ex] (g + 4.9)^2 = 11^2 - h^2 \\[3ex] (g + 4.9)^2 = 121 - 39.99 \\[3ex] (g + 4.9)^2 = 81.01 \\[3ex] g + 4.9 = \sqrt{81.01} \\[3ex] g + 4.9 = 9.000555538 \\[3ex] g = 9.000555538 - 4.9 \\[3ex] g = 4.100555538\;m $
(64.) NZQA An extra support structure is necessary for heavy winds.
The straight support beam will attach from M to a point Y, somewhere along PN so that the angle between the lines MY and PN will be 90°.
Calculate the perimeter of the triangle PYM.
Show your working clearly.


Number 64

$ Let\;\; |PY| = r \\[3ex] Let\;\; |MY| = k \\[3ex] |PN| = |PY| + |YN| ... diagram \\[3ex] 8 = r + |YN| \\[3ex] r + |YN| = 8 \\[3ex] |YN| = 8 - r \\[3ex] \underline{\triangle PYM} \\[3ex] |PM|^2 = |PY|^2 + |MY|^2 ...Pythagorean\;\;Theorem \\[3ex] 4.9^2 = r^2 + k^2 \\[3ex] r^2 + k^2 = 4.9^2 \\[3ex] k^2 = 4.9^2 - r^2...eqn.(1) \\[3ex] \underline{\triangle MYN} \\[3ex] |MN|^2 = |MY|^2 + |YN|^2 ...Pythagorean\;\;Theorem \\[3ex] h^2 = k^2 + (8 - r)^2 \\[3ex] k^2 + (8 - r)^2 = h^2 \\[3ex] k^2 = h^2 - (8 - r)^2 ...eqn.(2) \\[3ex] k^2 = k^2 \implies eqn.(2) = eqn.(1) \\[3ex] h^2 - (8 - r)^2 = 4.9^2 - r^2 \\[3ex] h^2 - [(8 - r)(8 - r)] = 24.01 - r^2 \\[3ex] 39.99 - (64 - 8r - 8r + r^2) = 24.01 - r^2 \\[3ex] 39.99 - (64 - 16r + r^2) = 24.01 - r^2 \\[3ex] 39.99 - 64 + 16r - r^2 = 24.01 - r^2 \\[3ex] -24.01 + 16r - r^2 = 24.01 - r^2 \\[3ex] 16r = 24.01 - r^2 + 24.01 + r^2 \\[3ex] 16r = 48.02 \\[3ex] r = \dfrac{48.02}{16} \\[5ex] r = 3.00125\;m \\[3ex] From\;\;eqn.(1) \\[3ex] k^2 = 4.9^2 - r^2 \\[3ex] k^2 = 4.9^2 - 3.00125^2...eqn.(1) \\[3ex] k^2 = 24.01 - 9.007501563 \\[3ex] k^2 = 15.00249844 \\[3ex] k = \sqrt{15.00249844} \\[3ex] k = 3.87330588\;m \\[3ex] \underline{\triangle PYM} \\[3ex] Perimeter \\[3ex] = |PY| + |MY| + |PM| \\[3ex] = r + k + 4.9 \\[3ex] = 3.00125 + 3.87330588 + 4.9 \\[3ex] = 11.77455588\;m $
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