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Solved Examples on Triangles

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 a wrong answer.

For SAT Students
Any question labeled SAT-C is a question that allows a calculator.
Any question labeled SAT-NC is a question that does not allow a calculator.

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.)

$ 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

$ \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] $

$ \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 $\triangle ABC$?

$ 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.

$ |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 $\overline{BP}$?

(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 $\triangle XYZ$, $\dfrac{\sin \angle X}{YZ} = \dfrac{\sin \angle Y}{XZ} = \dfrac{\sin \angle Z}{XY}$)

$ 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 $c^2 = a^2 + b^2 - 2ab \cos \angle C$

$ 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.) NZQA Angle EFC = 82°.
Angle FCD = 134°.
Lines BG and CF are parallel.
Lines AGFE and ABCD are both straight.

Find the size, z, of angle GAB.
Justify your answer.

$ \angle AFC + \angle EFC = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle AFC + 82 = 180 \\[3ex] \angle AFC = 180 - 82 \\[3ex] \angle AFC = 98^\circ \\[3ex] \angle ACF + \angle FCD = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle ACF + 134 = 180 \\[3ex] \angle ACF = 180 - 134 \\[3ex] \angle ACF = 46^\circ \\[3ex] \angle FAC = \angle GAB = z^\circ ...diagram \\[3ex] \underline{\triangle AFC} \\[3ex] \angle FAC + \angle AFC + \angle ACF = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle FAC + 98 + 46 = 180 \\[3ex] \angle FAC + 144 = 180 \\[3ex] \angle FAC = 180 - 144 \\[3ex] \angle FAC = 36^\circ $

(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

(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.

$ \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 $\angle BCD$?

$ 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 $\angle ACD$?

$ 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 $

Top

(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.)

$ (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

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.)

$ 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] $

$ \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} $

(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°

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

$ \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

$ \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 $\theta$

$ (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 On the diagram below, not drawn to scale, RQRQ = 9 m, RS = 12 m, ST = 13 m, ∠QRS = 60° and ∠SQT = 40°.

Calculate, correct to 1 decimal place,
(i.) the length QS
(ii.) the measure of ∠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] $

$ \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 $(x - 8)^\circ$

(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.) NZQA The diagram below represents the upper section of a crane.
Angle ABC = 27°.
Angle ADC = 35°.
AB = 8 metres.
Angle ACB = 90°.

Calculate the length, x, from B to D.
Show your working clearly.

$ \angle ABD = \angle ABC = 27^\circ ...diagram \\[3ex] \angle ADB = \angle ADC = 35^\circ ...diagram \\[3ex] \underline{\triangle BAD} \\[3ex] \angle BAD + \angle ABD + \angle ADB = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle BAD + 27 + 35 = 180 \\[3ex] \angle BAD + 62 = 180 \\[3ex] \angle BAD = 180 - 62 \\[3ex] \angle BAD = 118^\circ \\[3ex] \dfrac{|BD|}{\sin \angle BAD} = \dfrac{|AB|}{\sin \angle ADB}...Sine\;\;Law \\[5ex] \dfrac{x}{\sin 118^\circ} = \dfrac{8}{\sin \angle 35^\circ} \\[5ex] x = \dfrac{8\sin 118^\circ}{\sin \angle 35^\circ} \\[5ex] x = \dfrac{8(0.882947593)}{0.573576436} \\[5ex] x = \dfrac{7.063580744}{0.573576436} \\[5ex] x = 12.31497722 \\[3ex] x \approx 12\;m $

(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?

(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.) NZQA Lines PQ and AC are parallel to each other.
PQ = 12 metres.
AC = 21 metres.
AB = 25 metres.

Find the length of BP.
Show your working clearly.

$ \dfrac{|BP|}{|BA|} = \dfrac{|PQ|}{|AC|}...AA\sim\;\;Postulate \\[5ex] \dfrac{|BP|}{25} = \dfrac{12}{21} \\[5ex] |BP| = \dfrac{25 \cdot 12}{21} \\[5ex] |BP| = \dfrac{300}{21} \\[5ex] |BP| = 14.2857143 \\[3ex] |BP| \approx 14\;m $

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

$ 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 $

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

$ \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 $\angle x$?

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

$ \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 $

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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?

$ 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 $\angle C$?

$ 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°

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 $\triangle ABC$?

$ 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} $

(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 $\triangle PQR$?

$ 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 $

(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?

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

$ 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 $

(53.) NZQA (i) Part of the winding mechanism of a crane is shown below.

The shape of the winding mechanism is a regular pentagon, with each outside length measuring 12 metres from corner to corner, as shown above.
The winding mechanism has 5 metal arms.
Each arm is attached to the centre of the mechanism and to the corner of the pentagon.
What is the total length of all five metal arms?
Show your working clearly.

(ii) When the winding mechanism is in the shape of a regular n-sided polygon, each outside length measures 2z metres from corner to corner.
Calculate the total length of all the metal arms.
Give your answer in terms of n and z.
Show your working clearly.

$ (i) \\[3ex] Sum\;\;of\;\;the\;\;interior\;\;\angle s\;\;of\;\;a\;\;regular\;\;pentagon \\[3ex] = (5 - 2) * 180 \\[3ex] = 3 * 180 \\[3ex] = 540^\circ \\[3ex] Each\;\;interior\;\;\angle \\[3ex] = \dfrac{540}{5} \\[5ex] = 108^\circ \\[3ex] \implies \\[3ex] \angle A = \angle B = \angle C = \angle D = \angle E = 108^\circ \\[3ex] \angle OAB = \angle OBA = \angle OBC = \angle OCB = \angle OCD = \angle ODC = \\[3ex] = \angle ODE = \angle OED = \angle OEA = \angle OAE = \dfrac{108}{2} = 54^\circ ...base\;\;\angle s\;\;of\;\;isosceles\;\;\triangle s \\[5ex] \implies \\[3ex] \angle AOB = \angle BOC = \angle COD = \angle DOE = \angle EOA = p...other\;\;\angle \;\;of\;\;same\;\;isosceles\;\;\triangle s \\[3ex] p + 54 + 54 = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;each\;\;isosceles\;\;\triangle \\[3ex] p + 108 = 180 \\[3ex] p = 180 - 108 \\[3ex] p = 72 \\[3ex] \implies \\[3ex] \angle AOB = \angle BOC = \angle COD = \angle DOE = \angle EOA = 72^\circ \\[3ex] \underline{For\;\;each\;\;isosceles\;\;\triangle; \;\;say\;\;\triangle AOB} \\[3ex] \dfrac{|OB|}{\sin \angle OAB} = \dfrac{|AB|}{\sin \angle AOB}...Sine\;\;Law \\[5ex] \dfrac{|OB|}{\sin 54^\circ} = \dfrac{12}{\sin 72^\circ} \\[5ex] |OB| = \dfrac{12\sin 54^\circ}{\sin 72^\circ} \\[5ex] |OB| = \dfrac{12(0.8090169944)}{0.9510565163} \\[5ex] |OB| = \dfrac{9.708203932}{0.9510565163} \\[5ex] length\;\;of\;\;one\;\;metal\;\;arm = |OB| = 10.2078097 \\[3ex] \therefore total\;\;length\;\;of\;\;all\;\;five\;\;metal\;\;arms \\[3ex] = 5(10.2078097) \\[3ex] = 51.0390485 \\[3ex] \approx 51\;metres \\[3ex] $ (ii)
When the winding mechanism is in the shape of a regular pentagon, each outside length measures 12 metres from corner to corner.
Let us derive the total length of all the metal arms in terms of n and z

$ pentagon:\;\; n = 5 \\[3ex] 2z = 12 \\[3ex] 12 = 2z \\[3ex] 540 = (5 - 2) * 180 \\[3ex] 540 = 180(n - 2) \\[3ex] \dfrac{540}{5} = \dfrac{180(n - 2)}{5} \\[5ex] 108 = \dfrac{180(n - 2)}{n} \\[5ex] \dfrac{108}{2} = \dfrac{180(n - 2)}{2n} \\[5ex] 54 = \dfrac{90(n - 2)}{n} \\[5ex] 72 = 180 - 108 \\[3ex] 72 = 180 - \dfrac{180(n - 2)}{n} \\[5ex] 72 = \dfrac{180n}{n} - \dfrac{180(n - 2)}{n} \\[5ex] 72 = \dfrac{180n - 180(n - 2)}{n} \\[5ex] 72 = \dfrac{180n - 180n + 360}{n} \\[5ex] 72 = \dfrac{360}{n} \\[5ex] \implies \\[3ex] Length\;\;of\;\;one\;\;metal\;\;arm \\[3ex] = \dfrac{12\sin 54^\circ}{\sin 72^\circ} \\[5ex] = \dfrac{2z\sin \left[\dfrac{90(n - 2)}{n}\right]}{\sin \left(\dfrac{360}{n}\right)} \\[9ex] Total\;\;length\;\;of\;\;all\;\;five\;\;metal\;\;arms \\[3ex] = 5 * \dfrac{12\sin 54^\circ}{\sin 72^\circ} \\[5ex] = n * \dfrac{2z\sin \left[\dfrac{90(n - 2)}{n}\right]}{\sin \left(\dfrac{360}{n}\right)} \\[9ex] = \dfrac{2nz\sin \left[\dfrac{90(n - 2)}{n}\right]}{\sin \left(\dfrac{360}{n}\right)} $

(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 $\angle ACD$?

$ 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 $

(58.) ACT The degree measures of the 3 angles of the triangle below are expressed in terms of x
What is the value of x?

$ F.\;\; 10 \\[3ex] G.\;\; 12 \\[3ex] H.\;\; 24 \\[3ex] J.\;\; 30 \\[3ex] K.\;\; 36 \\[3ex] $

$ 4x + 5x + 6x = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] 15x = 180 \\[3ex] x = \dfrac{180}{15} \\[5ex] x = 12^\circ $

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.

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°

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°.

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.

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.

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.

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.

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.

$ 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 $