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 Bearings and Distances

Pre-requisites:
(1.) Angles
(2.) Bearings and Distances
(3.) Triangles

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.
Draw diagrams as applicable.
Show all work.

(1.) JAMB

From the diagram above, find the bearing of R from $S$

$A.\;\; 226^\circ \\[3ex] B.\;\; 224^\circ \\[3ex] C.\;\; 136^\circ \\[3ex] D.\;\; 134^\circ \\[3ex]$

$y = 180^\circ ...2\;\;right\;\;\angle s \\[3ex] x = 44^\circ ...alternate\;\; \angle s \\[3ex] Bearing\;\;of\;\;R\;\;from\;\;S \\[3ex] = y + p \\[3ex] = 180 + 44 \\[3ex] = 224^\circ$
(2.) ACT Juanita walked from her home to the bakery, first walking 0.3 miles due east and then 0.4 miles due north.
What is the straight-line distance, in miles, from the bakery to Juanita's home?

$A.\;\; 0.1 \\[3ex] B.\;\; 0.2 \\[3ex] C.\;\; 0.3 \\[3ex] D.\;\; 0.5 \\[3ex] E.\;\; 0.7 \\[3ex]$

Let us represent this information in a diagram

$Straight-line\;\;distance = hyp \\[3ex] hyp^2 = 0.3^2 + 0.4^2 ... Pythagorean\:\: Theorem \\[3ex] hyp^2 = 0.09 + 0.16 \\[3ex] hyp^2 = 0.25 \\[3ex] hyp = \sqrt{0.25} \\[3ex] hyp = 0.5\;miles$
(3.) JAMB A man walks 100 meters due West from a point X to Y
He then walks 100 meters due North to a point Z
Find the bearing of X from Z

$A.\;\; 195^\circ \\[3ex] B.\;\; 135^\circ \\[3ex] C.\;\; 225^\circ \\[3ex] D.\;\; 045^\circ \\[3ex]$

$Considering \triangle PQR \\[3ex] |XY| = |YZ| = 100\;m...equidistant \\[3ex] \angle YZX = \angle YXZ = p ...base\;\;\angle s\;\;of\;\;isosceles\triangle \\[3ex] \angle YZX + \angle YXZ + \angle XYZ = 180...sum\;\;of\;\;\angle s\;\;of\;\;\triangle PQR \\[3ex] \angle XYZ = 90 ...right\;\;\angle \\[3ex] p + p + 90 = 180 \\[3ex] 2p = 180 - 90 \\[3ex] 2p = 90 \\[3ex] p = \dfrac{90}{2} \\[5ex] p = 45^\circ \\[3ex] Bearing\;\;of\;\;P\;\;from\;\;R...red\;\;colored \\[3ex] = 180 - p...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] = 180 - 45 \\[3ex] = 135^\circ$
(4.) JAMB An aeroplane flies due north from airports P to Q and then flies due east to R
If Q is equidistant from P and R, find the bearing of P from R

$A.\;\; 225^\circ \\[3ex] B.\;\; 090^\circ \\[3ex] C.\;\; 270^\circ \\[3ex] D.\;\; 135^\circ \\[3ex]$

$Considering\;\; \triangle PQR \\[3ex] |PQ| = |QR|...equidistant \\[3ex] \angle QPR = \angle QRP = k ...base\;\;\angle s\;\;of\;\;isosceles\triangle \\[3ex] \angle QPR + \angle QRP + \angle PQR = 180...sum\;\;of\;\;\angle s\;\;of\;\;\triangle PQR \\[3ex] \angle PQR = 90 ...right\;\;\angle \\[3ex] k + k + 90 = 180 \\[3ex] 2k = 180 - 90 \\[3ex] 2k = 90 \\[3ex] k = \dfrac{90}{2} \\[5ex] k = 45^\circ \\[3ex] k = k_1 = 45^\circ ...alternate \angle s \\[3ex] Bearing\;\;of\;\;P\;\;from\;\;R \\[3ex] = 180 + k_1 \\[3ex] = 180 + 45 \\[3ex] = 225^\circ$
(5.) CSEC From a harbour, H, the bearing of two buoys, S and Q, are 185° and 311° respectively.
Q is 5.4 km from H while S is 3.5 km from H
(i) On the diagram below, which shows the sketch of this information, insert the value of the marked angle, QHS
(ii) Calculate QS, the distance between the two buoys.
(iii) Calculate the bearing of S from Q

$(i) \\[3ex] \angle QHS = 311 - 185 = 216^\circ \\[3ex]$

$(ii) \\[3ex] Let\;\; QS = h \\[3ex] h^2 = 5.4^2 + 3.5^2 - 2(5.4)(3.5) \cos 126^\circ...Cosine\;\;Law \\[3ex] h^2 = 29.16 + 12.25 - 37.8 * -0.5877852523 \\[3ex] h^2 = 41.41 + 22.21828254 \\[3ex] h^2 = 63.62828254 \\[3ex] h = \sqrt{63.62828254} \\[3ex] h = 7.976733826 \\[3ex] QS \approx 8.0\;km \\[3ex]$

$k + 126 + 185 = 360...\angle s\;\;at\;\;a\;\;point \\[3ex] k + 311 = 360 \\[3ex] k = 360 - 311 \\[3ex] k = 49^\circ \\[3ex] p = p ...alternate\;\; \angle s \\[3ex] p + k = 90 ...right\;\; \angle \\[3ex] p = 90 - k \\[3ex] p = 90 - 49 \\[3ex] p = 41^\circ \\[3ex] Considering\;\; \triangle QHS \\[3ex] \dfrac{\sin \angle SQH}{3.5} = \dfrac{\sin 126}{h}...Sine\;\;Law \\[5ex] \dfrac{\sin \angle SQH}{3.5} = \dfrac{0.8090169944}{7.976733826} \\[5ex] \sin \angle SQH = \dfrac{3.5 * 0.8090169944}{7.976733826} \\[5ex] \sin \angle SQH = \dfrac{2.83155948}{7.976733826} \\[5ex] \sin \angle SQH = 0.3549773055 \\[3ex] \angle SQH = \sin^{-1}(0.3549773055) \\[3ex] \angle SQH = 20.79205349 \\[3ex] (iii) \\[3ex] Bearing\;\;of\;\; S\;\;from\;\;Q \\[3ex] = 90 + p + \angle SQH \\[3ex] = 90 + 41 + 20.79205349 \\[3ex] = 151.7920535 \\[3ex] \approx 152^\circ$
(6.) CSEC The diagram below, not drawn to scale, shows the route of a ship cruising from Palmcity (P) to Quayton (Q) and then to Rivertown (R).
The bearing of Q from P is $133^\circ$ and the angle PQR is $56^\circ$

(i) Calculate the value of angle w
(ii) Determine the bearing of P from Q
(iii) Calculate the distance RP

$(i) \\[3ex] 56 + w = 133 ...alternate \angle s \;\;are\;\;equal \\[3ex] w = 133 - 56 \\[3ex] w = 77^\circ \\[3ex] (ii) \\[3ex] Bearing\;\;of\;\;P\;\;from\;\;Q \\[3ex] = 180 + w + 56 \\[3ex] = 180 + 133 \\[3ex] = 313^\circ \\[3ex] (iii) \\[3ex] RP^2 = 210^2 + 290^2 - 2(210)(290) \cos 56...Cosine\;\;Law \\[3ex] RP^2 = 44100 + 84100 - 121800 * 0.5591929035 \\[3ex] RP^2 = 128200 - 68109.69564 \\[3ex] RP^2 = 60090.30436 \\[3ex] RP = \sqrt{60090.30436} \\[3ex] RP = 245.133238 \\[3ex] RP \approx 245\;km$
(7.) ACT Olivia, Ashton, and Jane are standing on a soccer field such that Olivia is 20 meters due west of Ashton and Jane is 40 meters due north of Ashton.
Their positions are at the vertices of a triangle.
Which of the following expressions gives the degree measure of the angle of the triangle at the vertex where Olivia is standing?

$F.\;\; \cos^{-1}\left(\dfrac{40}{20}\right) \\[5ex] G.\;\; \sin^{-1}\left(\dfrac{40}{20}\right) \\[5ex] H.\;\; \sin^{-1}\left(\dfrac{20}{40}\right) \\[5ex] J.\;\; \tan^{-1}\left(\dfrac{40}{20}\right) \\[5ex] K.\;\; \tan^{-1}\left(\dfrac{20}{40}\right) \\[5ex]$

Let the degree measure of the angle of the triangle at the vertex where Olivia is standing be $\theta$
Let us represent the information using a diagram

$SOH:CAH:TOA \\[3ex] \tan \theta = \dfrac{40}{20} \\[5ex] \theta = \tan^{-1}\left(\dfrac{40}{20}\right)$
(8.) CSEC A ship travels from Akron (A) on a bearing of $030^\circ$ to Bellville (B), 90 km away.
It then travels to Comptin (C) which is 310 km due east of Akron (A), as shown in the diagram below.

(i.) Indicate on the diagram the bearing $030^\circ$ and the distances 90 km and 310 km
(ii.) Calculate, to the nearest km, the distance between Bellville (B) and Comptin (C)
(iii.) Calculate, to the nearest degree, the measure of $A\hat{B}C$
(iv.) Determine the bearing of Comptin (C) from Bellville (B)

(i.)

$(ii.) \\[3ex] \underline{\triangle ABC} \\[3ex] B\hat{A}C + 30 = 90 ...perpendicular\;\;angle \\[3ex] B\hat{A}C = 90 - 30 = 60^\circ \\[3ex] |BC|^2 = |AB|^2 + |AC|^2 - 2(|AB|)(|AC|) \cos B\hat{A}C \\[3ex] |BC|^2 = 90^2 + 310^2 - 2(90)(310) * \cos 60 \\[3ex] |BC|^2 = 8100 + 96100 - 55800 * 0.5 \\[3ex] |BC|^2 = 104200 - 27900 \\[3ex] |BC|^2 = 76300 \\[3ex] |BC| = \sqrt{76300} \\[3ex] |BC| = 276.2245463 \\[3ex] |BC| \approx 276\;km \\[3ex] (iii.) \\[3ex] \underline{\triangle ABC} \\[3ex] \dfrac{\sin A\hat{B}C}{310} = \dfrac{\sin B\hat{A}C}{|BC|}...Sine\;\;Law \\[5ex] \dfrac{\sin A\hat{B}C}{310} = \dfrac{\sin 60}{276.2245463} \\[5ex] \sin A\hat{B}C = \dfrac{310 * \sin 60}{276.2245463} \\[5ex] \sin A\hat{B}C = \dfrac{310 * 0.8660254038}{276.2245463} \\[5ex] \sin A\hat{B}C = \dfrac{268.4678752}{276.2245463} \\[5ex] \sin A\hat{B}C = 0.9719189651 \\[3ex] A\hat{B}C = \sin^{-1}(0.9719189651) \\[3ex] A\hat{B}C = 76.38976123 \\[3ex] A\hat{B}C \approx 76^\circ \\[3ex] (iv.) \\[3ex]$

$Bearing\;\;of\;\;Comptin\;\;from\;\;Bellville = \theta \\[3ex] \phi = 30^\circ...alternate\;\; \angle s\;\;are\;\;equal \\[3ex] \phi + \tau = A\hat{B}C ...diagram \\[3ex] 30 + \tau = 76.38976123 \\[3ex] \tau = 76.38976123 - 30 \\[3ex] \tau = 46.38976123 \\[3ex] \theta + \tau = 180...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \theta = 180 - \tau \\[3ex] \theta = 180 - 46.38976123 \\[3ex] \theta = 133.6102388 \\[3ex] \theta \approx 133^\circ$
(9.) WASSCE A town $J$ is $20\;km$ from a lorry station, $K$ on a bearing $065^\circ$
Another town, $T$ is $8\;km$ from $K$ on a bearing $155^\circ$
Calculate:
(i.) to the nearest kilometer, the distance of $T$ from $J$
(ii.) to the nearest degree, the bearing of $T$ from $J$

Let us draw the diagram

$(i.) \\[3ex] Distance\;\;of\;\;T\;\;from\;\;J = k \\[3ex] \angle TKJ = 25 + 65 = 90^\circ \\[3ex] k^2 = j^2 + t^2 - 2jt \cos \angle TKJ...Cosine\;\;Law \\[3ex] k^2 = 8^2 + 20^2 - 2(8)(20) * \cos 90 \\[3ex] k^2 = 64 + 400 - 320 * 0 \\[3ex] k^2 = 464 - 0 \\[3ex] k^2 = 464 \\[3ex] k = \sqrt{464} \\[3ex] k = 21.54065923 \\[3ex] k \approx 22\;km \\[3ex] (ii.) \\[3ex] \dfrac{\sin \theta}{j} = \dfrac{\sin 90}{k}...Sine\;\;Law \\[5ex] \dfrac{\sin \theta}{8} = \dfrac{\sin 90}{21.54065923} \\[5ex] \sin \theta = \dfrac{8 * \sin 90}{21.54065923} \\[5ex] \sin \theta = \dfrac{8 * 1}{21.54065923} \\[5ex] \sin \theta = 0.3713906763 \\[3ex] \theta = \sin^{-1}{0.3713906763} \\[3ex] \theta = 21.80140948 \\[3ex] \theta + \phi = 65 ...alternate \angle s\;\;are\;\;equal \\[3ex] \phi = 65 - \theta \\[3ex] \phi = 65 - 21.80140948 \\[3ex] \phi = 43.19859052 \\[3ex] Bearing\;\;of\;\;T\;\;from\;\;J \\[3ex] = 180 + \phi \\[3ex] = 180 + 43.19859052 \\[3ex] = 223.1985905 \\[3ex] \approx 223^\circ$
(10.) curriculum.gov.mt What is the bearing of Q from P?

$k = k ...alternate \;\;\angle s\;\;are\;\;equal \\[3ex] k + 180 + 90 = 325 ...diagram \\[3ex] k + 270 = 325 \\[3ex] k = 325 - 270 \\[3ex] k = 55^\circ \\[3ex] Bearing\;\;of\;\;Q\;\;from\;\;P \\[3ex] = 90 + k \\[3ex] = 90 + 55 \\[3ex] = 145^\circ$
(11.) CSEC A boat leaves a dock at point A and travels for a distance of 15 km to point B on a bearing of $135^\circ$
The boat then changes course and travels for a distance of 8 km to point C on a bearing of $060^\circ$

(a.) Illustrate the above diagram in a clearly labelled diagram
The diagram should show the
(i.) north direction
(ii.) bearings $135^\circ$ and $060^\circ$
(iii) distances 8 km and 15 km

(b.) Calculate
(i.) the distance AC
(ii.) $\angle BCA$
(iii.) the bearing of A from C

(a.)

$(b.) \\[3ex] \theta = 45^\circ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \theta + \phi = 90...complementary\;\;\angle s \\[3ex] \phi = 90 - \theta \\[3ex] \phi = 90 - 45 \\[3ex] \phi = 45^\circ \\[3ex] \angle ABC = B = \phi + 60...diagram \\[3ex] B = 45 + 60 \\[3ex] B = 105^\circ \\[3ex] (i.) \\[3ex] Distance\;\; AC = b \\[3ex] b^2 = a^2 + c^2 - 2ac \cos B...Cosine\;\;Law \\[3ex] b^2 = 8^2 + 15^2 - 2(8)(15) * \cos 105 \\[3ex] b^2 = 64 + 225 - 240 * - 0.2588190451 \\[3ex] b^2 = 289 + 62.11657082 \\[3ex] b^2 = 351.1165708 \\[3ex] b = \sqrt{351.1165708} \\[3ex] b = 18.73810478 \\[3ex] b \approx 19\;km \\[3ex] (ii.) \\[3ex] \angle BCA = \angle C \\[3ex] \dfrac{\sin C}{c} = \dfrac{\sin B}{b} \\[5ex] \dfrac{\sin C}{15} = \dfrac{\sin 105}{18.73810478} \\[5ex] \sin C = \dfrac{15\sin 105}{18.73810478} \\[5ex] \sin C = \dfrac{15(0.9659258263)}{18.73810478} \\[5ex] \sin C = \dfrac{14.48888739}{18.73810478} \\[5ex] \sin C = 0.773231208 \\[3ex] C = \sin^{-1}(0.773231208) \\[3ex] C = 50.64494187 \\[3ex] C \approx 51^\circ \\[3ex]$

$(iii) \\[3ex] \tau = 60^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] Bearing\;\;of\;\;A\;\;from\;\;C \\[3ex] = 180 + \tau + \angle BCA \\[3ex] = 180 + 60 + 50.64494187 \\[3ex] = 290.6449419 \\[3ex] \approx 291^\circ$
(12.) ACT Two motorcycles, starting at the same point at the same time, travel away from each other at a 90° angle.
One travels at 40 miles per hour and the other at 60 miles per hour.
If they continue traveling at these constant rates, after about how many hours will they be 200 miles apart?

$A.\;\; 1.4 \\[3ex] B.\;\; 2.8 \\[3ex] C.\;\; 3.2 \\[3ex] D.\;\; 7.7 \\[3ex] E.\;\; 8.7 \\[3ex]$

The diagram representing the information is shown:

$e^2 = 60^2 + 40^2 ...Pythagorean\;\;Theorem \\[3ex] e^2 = 3600 + 1600 \\[3ex] e^2 = 5200 \\[3ex] e = \sqrt{5200} \\[3ex] e = 72.11102551\;mph \\[3ex] But: \\[3ex] time = \dfrac{distance}{speed} \\[5ex] distance = 200\;miles \\[3ex] speed = 72.11102551\;mph \\[3ex] time = \dfrac{200}{72.11102551} \\[5ex] time = 2.773500981 \\[3ex] time \approx 2.8\;seconds$
(13.) curriculum.gov.mt Three flag poles P, Q and R are fixed to the ground on a flat field.
The bearing of P from Q is 243°
The bearing of R from Q is 153°
The distance PR is 67 m and the distance QR is 54 m.

(a) Show that $P\hat{Q}R = 90^\circ$
(b) Calculate $Q\hat{P}R$
(c) Calculate the bearing of R from P

Indicate the bearings on the triangle

$(a) \\[3ex] Bearing\;\;of\;\;P\;\;from\;\;Q = 243^\circ \\[3ex] Bearing\;\;of\;\;R\;\;from\;\;Q = 153^\circ \\[3ex] P\hat{Q}R = 243 - 153 ...diagram \\[3ex] P\hat{Q}R = 90^\circ \\[3ex] (b) \\[3ex] \dfrac{\sin Q\hat{P}R}{QR} = \dfrac{\sin P\hat{Q}R}{PR} ...Sine\;\;Law \\[5ex] \dfrac{\sin Q\hat{P}R}{54} = \dfrac{\sin 90}{67} \\[5ex] \sin Q\hat{P}R = \dfrac{54\sin 90}{67} \\[5ex] \sin Q\hat{P}R = \dfrac{54(1)}{67} \\[5ex] \sin Q\hat{P}R = \dfrac{54}{67} \\[5ex] \sin Q\hat{P}R = 0.8059701493 \\[3ex] Q\hat{P}R = \sin^{-1}(0.8059701493) \\[3ex] Q\hat{P}R = 53.70405214^\circ \\[3ex]$

$(c) \\[3ex] k + 243 = 270^\circ...diagram \\[3ex] k = 270 - 243 \\[3ex] k = 27^\circ \\[3ex] k + x = 90^\circ ...diagram \\[3ex] 27 + x = 90 \\[3ex] x = 90 - 27 \\[3ex] x = 63^\circ \\[3ex] x = x ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] Bearing\;\;of\;\;R\;\;from\;\;P \\[3ex] = x + Q\hat{P}R \\[3ex] = 63 + 53.70405214 \\[3ex] = 116.7040521^\circ$
(14.) WASSCE A bearing of 320° expressed as a compass bearing is

$A.\;\; N\;50^\circ\;W \\[3ex] B.\;\; N\;40^\circ\;W \\[3ex] C.\;\; N\;50^\circ\;E \\[3ex] D.\;\; N\;40^\circ\;E \\[5ex]$

$320 + \phi = 360 ...\angle s\;\;at\;\;a\;\;point \\[3ex] \phi = 360 - 320 \\[3ex] \phi = 40^\circ \\[3ex] Compass\;\;Bearing = N\;40^\circ\;W$
(15.) NZQA A spider is crawling along level ground.
The spider starts at point S and crawls directly north for a distance of 54 cm, until it reaches point H.
The spider then changes direction and heads to point F, which is 140 cm away, on a bearing of 078°
SH = 54 cm        HF = 140 cm

Find the direct distance and bearing of S from F.

$\underline{Direct\;\;Distance\;\;of\;\;S\;\;from\;\;F} \\[3ex] \angle SHF + 78 = 180^\circ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle SHF = 180 - 78 \\[3ex] \angle SHF = 102^\circ \\[3ex] \underline{\triangle SHF} \\[3ex] |SF|^2 = |SH|^2 + |HF|^2 - 2 * |SH| * |HF| * \cos \angle SHF...Cosine\;\;Law \\[3ex] |SF|^2 = 54^2 + 140^2 - 2(54)(140) * \cos 102 \\[3ex] |SF|^2 = 2916 + 19600 - 15120 * -0.2079116908 \\[3ex] |SF|^2 = 22516 + 3143.624765 \\[3ex] |SF|^2 = 25659.62477 \\[3ex] |SF| = \sqrt{25659.62477} \\[3ex] |SF| = 160.186219\;cm \\[3ex] |SF| \approx 160\;cm \\[3ex]$

$\dfrac{\sin \angle HSF}{|HF|} = \dfrac{\sin \angle SHF}{|SF|}...Sine\;\;Law \\[5ex] \dfrac{\sin \angle HSF}{140} = \dfrac{\sin 102}{160.186219} \\[5ex] \sin \angle HSF = \dfrac{140 \sin 102}{160.186219} \\[5ex] \sin \angle HSF = \dfrac{140(0.9781476007)}{160.186219} \\[5ex] \sin \angle HSF = \dfrac{136.9406641}{160.186219} \\[5ex] \sin \angle HSF = 0.8548841777 \\[3ex] \angle HSF = \sin^{-1}(0.8548841777) \\[3ex] \angle HSF = 58.74694159^\circ \\[3ex] Let\;\;\angle HSF = k \\[3ex] k = k = 58.74694159^\circ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{Direct\;\;Bearing\;\;of\;\;S\;\;from\;\;F} \\[3ex] Bearing \\[3ex] = 180 + k \\[3ex] = 180 + 58.74694159 \\[3ex] = 238.7469416^\circ$
(16.) JAMB The bearing of a bird on a tree from a hunter on the ground is $N\;72^\circ\;E$
What is the bearing of the hunter from the bird?

$A.\;\; S\;18^\circ\;W \\[3ex] B.\;\; S\;72^\circ\;W \\[3ex] C.\;\; S\;72^\circ\;E \\[3ex] D.\;\; S\;27^\circ\;E \\[3ex] E.\;\; S\;27^\circ\;W \\[3ex]$

$\theta = 72^\circ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] Bearing\;\;of\;\;the\;\;hunter\;\;from\;\;the\;\;bird \\[3ex] = 180 + \theta \\[3ex] = 180 + 72 \\[3ex] = 252^\circ \\[3ex] OR \\[3ex] Compass\;\;Bearing\;\;of\;\;the\;\;hunter\;\;from\;\;the\;\;bird \\[3ex] = S\;\theta\;W \\[3ex] = S\;72^\circ\;W$
(17.) CSEC (i) Draw a diagram to represent the information given below.
Show clearly the north line in your diagram
Town F is 50 km east of town G
Town H is on a bearing of 040° from town F
The distance from F to H is 65 km

(ii) Calculate, to the nearest kilometre, the actual distance GH
(iii) Calculate, to the nearest degree, the bearing of H from G

(i)

$(ii) \\[3ex] \angle GFH = 90 + 40 = 130^\circ \\[3ex] |GH|^2 = |GF|^2 + |FH|^2 - 2 * |GF| * |FH| * \cos \angle GFH...Cosine\;\;Law \\[3ex] |GH|^2 = 50^2 + 65^2 - 2(50)(65) * \cos 130 \\[3ex] |GH|^2 = 2500 + 4225 - 6500 * -0.6427876097 \\[3ex] |GH|^2 = 6725 + 4178.119463 \\[3ex] |GH|^2 = 10903.11946 \\[3ex] |GH| = \sqrt{10903.11946} \\[3ex] |GH| = 104.4180035 \\[3ex] |GH| \approx 104\;km \\[3ex] (iii) \\[3ex] \dfrac{\sin \angle HGF}{|FH|} = \dfrac{\sin \angle GFH}{|GH|}...Sine\;\;Law \\[5ex] \dfrac{\sin \phi}{65} = \dfrac{\sin 130}{104.4180035} \\[5ex] \sin \phi = \dfrac{65 \sin 130}{104.4180035} \\[5ex] \sin \phi = \dfrac{65 * 0.7660444431}{104.4180035} \\[5ex] \sin \phi = \dfrac{49.7928888}{104.4180035} \\[5ex] \sin \phi = 0.4768611459 \\[3ex] \phi = \sin^{-1}(0.4768611459) \\[3ex] \phi = 28.48059838 \\[3ex] \theta + \phi = 90 ...right\;\;\angle \\[3ex] \theta = 90 - \phi \\[3ex] \theta = 90 - 28.48059838 \\[3ex] \theta = 61.51940162 \\[3ex] Bearing\;\;of\;\;H\;\;from\;\;G = \theta \approx 62^\circ$
(18.)

(19.) ACT Loto begins at his back door and walks 8 yards east, 6 yards north, 12 yards east, and 5 yards north to the barn door.
About how many yards less would he walk if he could walk directly from the back door to the barn door?

$A.\;\; 8 \\[3ex] B.\;\; 19 \\[3ex] C.\;\; 23 \\[3ex] D.\;\; 26 \\[3ex] E.\;\; 31 \\[3ex]$

Let us draw the diagram to represent the information.
Walking 8 yards east means walking 8 yards due/directly east
This question is trying to assess our knowledge of the Pythagorean Theorem, as well as the importance of the hypotenuse compared to the two legs of a right triangle.

$c^2 = 6^2 + 8^2 ...Pythagorean\;\;Theorem \\[3ex] c^2 = 36 + 64 \\[3ex] c^2 = 100 \\[3ex] c = \sqrt{100} \\[3ex] c = 10\;yards \\[3ex] Also: \\[3ex] d^2 = 5^2 + 12^2 ...Pythagorean\;\;Theorem \\[3ex] d^2 = 25 + 144 \\[3ex] d^2 = 169 \\[3ex] d = \sqrt{169} \\[3ex] d = 13\;yards \\[3ex] \underline{Distance\;\;walked\;\;by\;\;Loto} \\[3ex] = 8 + 6 + 12 + 5 \\[3ex] = 31\;yards \\[3ex] \underline{Distance\;\;that\;\;Loto\;\;could\;\;have\;\;walked\;\;directly} \\[3ex] = c + d \\[3ex] = 10 + 13 \\[3ex] = 23\;yards \\[3ex] \underline{How\;\;many\;\;yards\;\;less} \\[3ex] = Difference \\[3ex] = 31 - 23 \\[3ex] = 8\;yards$
(20.) JAMB From a point Z, 60m north of X, a man walks 60√3 m eastwards to another point Y.
Find the bearing of Y from X

$A.\;\; 030^\circ \\[3ex] B.\;\; 045^\circ \\[3ex] C.\;\; 060^\circ \\[3ex] D.\;\;090^\circ \\[3ex]$

Let us represent the information on a diagram

$\tan \theta = \dfrac{opp}{adj}...SOHCAHTOA \\[5ex] \tan \theta = \dfrac{60\sqrt{3}}{60} \\[5ex] \tan\theta = \sqrt{3} \\[3ex] \theta = \tan^{-1}(\sqrt{3}) \\[3ex] \theta = 60^\circ$

(21.) NZQA Captain Cook's ship, HMS Endeavour, sailed along the coastline via the route shown below on the map of the North Island, New Zealand.
The ship sailed from G to S on a bearing of 054° for a distance of 448 km.
It then changed direction, sailing from S to M on a bearing of 294° for a further distance of 635 km.

If Captain Cook was able to fly directly from G to M, find the bearing of M from G.

Let us represent the information in the diagram.

$\underline{\triangle GSM} \\[3ex] \angle GSM = 294 - (180 + 54) \\[3ex] \angle GSM = 294 - 234 \\[3ex] \angle GSM = 60^\circ \\[3ex] |MG|^2 = |MS|^2 + |GS|^2 - 2 * |MS| * |GS| * \cos \angle GSM ...Cosine\;\;Law \\[3ex] |MG|^2 = 635^2 + 448^2 - 2(635)(448) * \cos 60^\circ \\[3ex] |MG|^2 = 403225 + 200704 - 568960 * \dfrac{1}{2} \\[5ex] |MG|^2 = 603929 - 284480 \\[3ex] |MG|^2 = 319449 \\[3ex] |MG| = \sqrt{319449} \\[3ex] |MG| = 565.1981953\;km \\[3ex] \dfrac{\sin \angle MGS}{|MS|} = \dfrac{\sin \angle GSM}{|MG|} ... Sine\;\;Law \\[5ex] \dfrac{\sin \angle MGS}{635} = \dfrac{\sin 60^\circ}{565.1981953} \\[5ex] \sin \angle MGS = \dfrac{635 * \sin 60}{565.1981953} \\[5ex] \sin \angle MGS = \dfrac{635 * 0.8660254038}{565.1981953} \\[5ex] \sin \angle MGS = \dfrac{549.9261314}{565.1981953} \\[5ex] \sin \angle MGS = 0.972979277 \\[3ex] \angle MGS = \sin^{-1}(0.972979277) \\[3ex] \angle MGS = 76.65037991^\circ \\[3ex] Bearing\;\;of\;\;M\;\;from\;\;G \\[3ex] = 360 - (\angle MGS - 54) \\[3ex] = 360 - (76.65037991 - 54) \\[3ex] = 360 - 22.65037991 \\[3ex] = 337.3496201^\circ$
(22.)

(23.) CSEC From a harbour, H, the bearing of two ships, Q and R, are 069° and 151° respectively.
Q is 175 km from H while R is 242 km from H

(i) Complete the diagram above to show the information given.
(ii) Calculate QR, the distance between the two ships, to the nearest km.
(iii) Calculate how far due south is Ship R of the harbour, H

(i)

$(ii) \\[3ex] |QR|^2 = |QH|^2 + |RH|^2 - 2 * |QH| * |RH| * \cos \angle QHR...Cosine\;\;Law \\[3ex] |QR|^2 = 175^2 + 242^2 - 2(175)(242) * \cos 82 \\[3ex] |QR|^2 = 30625 + 58564 - 84700 * 0.139173101 \\[3ex] |QR|^2 = 89189 - 11787.96165 \\[3ex] |QR|^2 = 77401.03835 \\[3ex] |QR| = \sqrt{77401.03835} \\[3ex] |QR| = 278.210421 \\[3ex] |QR| \approx 278\;km \\[3ex]$

How far due south is Ship R of the harbour, H = |PH|

$69 + 82 + \theta = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 151 + \theta = 180 \\[3ex] \theta = 180 - 151 \\[3ex] \theta = 29^\circ \\[3ex] \underline{\triangle HPR} \\[3ex] \cos \theta = \dfrac{|PH|}{|RH|}...SOHCAHTOA \\[3ex] \cos 29 = \dfrac{|PH|}{242} \\[5ex] 242 \cos 29 = |PH| \\[3ex] |PH| = 242 * \cos 29 \\[3ex] |PH| = 242(0.8746197071) \\[3ex] |PH| = 211.6579691 \\[3ex] |PH| \approx 212\;km$
(24.) GCSE (a) Here is a map showing points A and B

Kemal wants to measure the bearing of A from B
He draws two lines and measures the angle between them.

Kemal says that the bearing of A from B is 100°
Is his method correct?

(b) On a different map, the bearing of D from C is 045°
Nina says,
"D is North West of C"
Is Nina correct?

(c) This map shows an airport, E, on an island.

A plane files due South from the airport.
How far does it fly until it reaches the sea?

(a) Kemal's method is not correct.
To get the bearing of A from B, he should begin from B, drawing from the North pole of B until it touches the line of A

(b)

Nina is not correct.
D is North East of C

(For learning purposes: C is South West of D)

(c) Use a ruler (in cm) to measure the vertical distance (because of due South) from airport Point E to the sea (diagram)
Multiply the length (in cm) by 100
(25.) JAMB A ship H leaves a port P and sails 30km due south.
Then it sails 60km due west.
What is the bearing of H from P?

$A.\;\; 26^\circ 34' \\[3ex] B.\;\; 243^\circ 26' \\[3ex] C.\;\; 116^\circ 34' \\[3ex] D.\;\; 63^\circ 26' \\[3ex] E.\;\; 240^\circ \\[3ex]$

Let us represent the information on a diagram

$\tan \theta = \dfrac{opp}{adj}...SOHCAHTOA \\[5ex] \tan \theta = \dfrac{60}{30} \\[5ex] \tan \theta = 2 \\[3ex] \theta = \tan^{-1}(2)...Acute\;\;\angle \\[3ex] Bearing\;\;of;\;H\;\;from\;\;P \\[3ex] = 180 + \theta \\[3ex] = 180 + \tan^{-1}(2)...Reflex\;\;\angle \\[3ex]$ Because JAMB does not require a calculator:
Options A, C, and D are incorrect because they are not reflex angles.
Option E is also incorrect because $\tan^{-1}(2)$ is a decimal
This leaves us with Option B as the correct option.
(26.)

(27.) CSEC From a port, L, ship R is 250 kilometres on a bearing of 065°
Ship T is 180 kilometres from L on a bearing of 148°
This information is illustrated in the diagram below.

(i) Complete the diagram above by inserting the value of angle RLT
(ii) Calculate RT, the distance between the two ships.
(iii) Determine the bearing of T from R.

The information is represented on the diagram:

$(i) \\[3ex] \angle RLT + 65^\circ = 148^\circ...bearing\;\;of\;\;T\;\;from\;\;L \\[3ex] \angle RLT = 148 - 65 \\[3ex] \angle RLT = 83^\circ \\[3ex] (ii) \\[3ex] |RT|^2 = |LT|^2 + |LR|^2 - 2 * |LT| * |LR| * \cos \angle RLT...Cosine\;\;Law \\[3ex] |RT|^2 = 180^2 + 250^2 - 2(180)(250) * \cos 83^\circ \\[3ex] |RT|^2 = 32400 + 62500 - (90000 * 0.1218693434) \\[3ex] |RT|^2 = 94900 - 10968.24091 \\[3ex] |RT|^2 = 83931.75909 \\[3ex] |RT| = \sqrt{83931.75909} \\[3ex] |RT| = 289.7097843 \\[3ex] |RT| \approx 290\;km \\[3ex]$

$\dfrac{\sin \angle LRT}{|LT|} = \dfrac{\sin \angle RLT}{|RT|} ...Sine\;\;Law \\[5ex] \dfrac{\sin \theta}{180} = \dfrac{\sin 83}{289.7097843} \\[5ex] \sin \theta = \dfrac{180 * \sin 83}{289.7097843} \\[5ex] \sin \theta = \dfrac{178.6583073}{289.7097843} \\[5ex] \sin \theta = 0.6166802676 \\[3ex] \theta = \sin^{-1}(0.6166802676) \\[3ex] \theta = 38.07411323^\circ \\[3ex] \psi + \theta = 65^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \psi = 65 - \theta \\[3ex] \psi = 65 - 38.07411323^\circ \\[3ex] \psi = 26.92588677^\circ \\[3ex] (iii) \\[3ex] Bearing\;\;of\;\;T\;\;from\;\;R \\[3ex] = 180 + \psi \\[3ex] = 180 + 26.92588677^\circ \\[3ex] = 206.9258868^\circ$
(28.)

(29.) WASSCE The bearing of points X and Y from Z are 040° and 300°, respectively.
If |XY| = 19.5km and |YZ| = 11.5km
a) Illustrate the information in a diagram
b) Calculate, correct to the nearest whole number,
i) ∠ZXY
ii) |XZ|

(a.) The diagram is:

$(b.) \\[3ex] (i.) \\[3ex] \dfrac{\sin \angle ZXY}{|ZY|} = \dfrac{\sin \angle YZX}{|YX|}...Sine\;\;Law \\[5ex] \dfrac{\sin \theta}{11.5} = \dfrac{\sin 100}{19.5} \\[5ex] \sin \theta = \dfrac{11.5 * \sin 100}{19.5} \\[5ex] \sin \theta = \dfrac{11.5 * 0.984807753}{19.5} \\[5ex] \sin \theta = 0.5807840595 \\[3ex] \theta = \sin^{-1}(0.5807840595) \\[3ex] \theta = 35.50570812 \\[3ex] \theta \approx 36^\circ \\[3ex] \angle ZYX + \angle YZX + \angle ZXY = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle ZYX + 100 + 35.50570812 = 180 \\[3ex] \angle ZYX = 180 - 100 - 35.50570812 \\[3ex] \angle ZYX = 44.49429188^\circ \\[3ex] (ii.) \\[3ex] \dfrac{|XZ|}{\sin \angle ZYX} = \dfrac{|YX|}{\sin \angle YZX}...Sine\;\;Law \\[5ex] \dfrac{|XZ|}{\sin 44.49429188} = \dfrac{19.5}{\sin 100} \\[5ex] |XZ| = \dfrac{19.5 * \sin 44.49429188}{\sin 100} \\[5ex] |XZ| = \dfrac{19.5 * 0.7008382029}{0.984807753} \\[5ex] |XZ| = 13.66634496 \\[3ex] |XZ| \approx 14\;km$
(30.)

(31.)

(32.)

(33.) CSEC The diagram below, not drawn to scale, shows the relative positions of three reservoirs B, F and G, all on level ground.
The distance BF = 32 km, FG = 55 km, ∠BFG is 103° and F is on a bearing of 042° from B.

(i) Determine the bearing of B from F.
(ii) Calculate the distance BG, giving your answer to one decimal place.
(iii) Calculate, to the nearest degree, the bearing of G from B.

(i)

$\phi = 42^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] Bearing\;\;of\;\;B\;\;from\;\;F\;\;(blue\;\;color) \\[3ex] = 180 + \phi \\[3ex] = 180 + 42 \\[3ex] = 222^\circ \\[3ex] (ii) \\[3ex] |BG|^2 = |BF|^2 + |FG|^2 - 2 * |BF| * |FG| * \cos \angle BFG ...Cosine\;\;Law \\[3ex] |BG|^2 = 32^2 + 55^2 - 2(32)(55) * \cos 103^\circ \\[3ex] |BG|^2 = 1024 + 3025 - 3520(-0.2249510543) \\[3ex] |BG|^2 = 4049 + 791.8277113 \\[3ex] |BG|^2 = 4840.827711 \\[3ex] |BG| = \sqrt{4840.827711} \\[3ex] |BG| = 69.57605703 \\[3ex] |BG| \approx 70\;km \\[3ex]$

$\dfrac{\sin \angle FBG}{|FG|} = \dfrac{\sin \angle BFG}{|BG|} ...Sine\;\;Law \\[5ex] \dfrac{\sin \angle FBG}{55} = \dfrac{\sin 103^\circ}{69.57605703} \\[5ex] \sin \angle FBG = \dfrac{55\sin 103^\circ}{69.57605703} \\[5ex] \sin \angle FBG = \dfrac{55(0.9743700648)}{69.57605703} \\[5ex] \sin \angle FBG = 0.7702413136 \\[3ex] \angle FBG = \sin^{-1}(0.7702413136) \\[3ex] \angle FBG = 50.37556355^\circ \\[3ex] Bearing\;\;of\;\;G\;\;from\;\;B\;\;(red\;\;color) \\[3ex] = 42 + \angle FBG \\[3ex] = 42 + 50.37556355 \\[3ex] = 92.37556355 \\[3ex] \approx 92^\circ$
(34.)

(35.)

(36.)

(37.) CSEC A ship leaves Port A and sails 52 km on a bearing of 044° to Port B.
The ship then changes course to sail to Port C, 72 km away, on a bearing of 105°.
(i) On the diagram below, not drawn to scale, label the known distances travelled and the known angles.

(ii) Determine the measure of ∠ABC.
(iii) Calculate, to the nearest km, the distance AC.
(iv) Show that the bearing of A from C, to the nearest degree, is 260°.

(i) The labelled diagram is:

(ii)
Let us indicate more labels to help us with the calculations.

$\phi = 44^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] 105 + \theta = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \theta = 180 - 105 \\[3ex] \theta = 75^\circ \\[3ex] (ii) \\[3ex] \angle ABC = \phi + \theta ...diagram \\[3ex] \angle ABC = 44 + 75 \\[3ex] \angle ABC = 119^\circ \\[3ex] (iii) \\[3ex] |AC|^2 = |AB|^2 + |BC|^2 - 2 * |AB| * |BC| * \cos \angle ABC...Cosine\;\;Law \\[3ex] |AC|^2 = 52^2 + 72^2 - 2(52)(72)(\cos 119^\circ) \\[3ex] |AC|^2 = 2704 + 5184 - 7488(-0.4848096202) \\[3ex] |AC|^2 = 7888 + 3630.254436 \\[3ex] |AC|^2 = 11518.25444 \\[3ex] |AC| = \sqrt{11518.25444} \\[3ex] |AC| = 107.3231309 \\[3ex] |AC| \approx 107\;km \\[3ex] (iv) \\[3ex] \dfrac{\sin \angle ACB}{|AB|} = \dfrac{\sin \angle ABC}{|AC|}...Sine\;\;Law \\[5ex] \dfrac{\sin \angle ACB}{52} = \dfrac{\sin 119^\circ}{107.3231309} \\[5ex] \sin \angle ACB = \dfrac{52\sin 119^\circ}{107.3231309} \\[5ex] \sin \angle ACB = \dfrac{52(0.8746197071)}{107.3231309} \\[5ex] \sin \angle ACB = 0.423769083 \\[3ex] \angle ACB = \sin^{-1}(0.423769083) \\[3ex] \angle ACB = 25.07277523^\circ \\[3ex] \beta = \theta = 75^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] Bearing\;\;of\;\;A\;\;from\;\;C...indicated\;\;by\;\;\color{purple}{purple\;\;color} \\[3ex] = 360 - (\beta + \angle ACB) ...\angle s\;\;at\;\;a\;\;point \\[3ex] = 360 - (75 + 25.07277523) \\[3ex] = 360 - 100.0727752 \\[3ex] = 259.9272248 \\[3ex] \approx 260^\circ$
(38.)

(39.) NZQA Three ships, K, L, and M, are floating on the surface of the sea, as shown in the diagram below.
The bearing of L from K is 126°.
The angle LKM = 111°.
KM = 4.4 km.
KL = 9.8 km.

Find the distance AND bearing of M from L.

$\underline{Distance\;\;of\;\;M\;\;from\;\;L = |LM|} \\[3ex] |LM|^2 = |KM|^2 + |KL|^2 - 2 * |KM| * |KL| * \cos \angle LKM...Cosine\;\;Law \\[3ex] |LM|^2 = (4.4)2 + (9.8)2 - 2(4.4)(9.8) * \cos 111^\circ \\[3ex] |LM|^2 = 19.36 + 96.04 - 86.24 * -0.3583679495 \\[3ex] |LM|^2 = 115.4 + 30.90565197 \\[3ex] |LM|^2 = 146.305652 \\[3ex] |LM| = \sqrt{146.305652} \\[3ex] |LM| = 12.096568733 \\[3ex] |LM| \approx 12.1\;km \\[3ex]$ To find the bearing of M from L, let us review the diagram

$\dfrac{\sin \angle KLM}{|KM|} = \dfrac{\sin \angle LKM}{|LM|} ...Sine\;\;Law \\[5ex] \dfrac{\sin\phi}{4.4} = \dfrac{\sin 111}{12.096568733} \\[5ex] \sin \phi = \dfrac{4.4 * \sin 111}{12.096568733} \\[5ex] \sin \phi = \dfrac{4.4(0.9335804265)}{12.096568733} \\[5ex] \sin \phi = 0.3395800881 \\[3ex] \phi = \sin^{-1}(0.3395800881) \\[3ex] \phi = 19.85129284^\circ \\[3ex] \theta + 90 = 126^\circ ... diagram \\[3ex] \theta = 126 - 90 \\[3ex] \theta = 36^\circ \\[3ex] \theta(1st) = \theta(2nd)...red\;\;color = 36^\circ...alternate\;\;\angle s \;\;are\;\;equal \\[3ex] \beta + \phi = \theta ...diagram \\[3ex] \beta + 19.85129284 = 36 \\[3ex] \beta = 36 - 19.85129284 \\[3ex] \beta = 16.14870716^\circ \\[3ex] \underline{Bearing\;\;of\;\;M\;\;from\;\;L = \psi} \\[3ex] \psi = 270 + \beta ...diagram \\[3ex] \psi = 270 + 16.14870716 \\[3ex] \psi = 286.1487072 \\[3ex] \psi \approx 286^\circ$
(40.)

(41.) CSEC The diagram below shows straight roads connecting the towns L, M, N and R.
LR = 18 km, LN = 12 km and MN = 10 km.
Angle RLN = 25° and angle LMN = 88°

(i) Calculate angle MLN.
(ii) Calculate the distance NR.
(iii) Determine the bearing of Town R from Town L.

$(i) \\[3ex] \underline{\triangle LMN} \\[3ex] \dfrac{\sin \angle MLN}{|MN|} = \dfrac{\sin \angle LMN}{|LN|}...Sine\;\;Law \\[5ex] \dfrac{\sin \angle MLN}{10} = \dfrac{\sin 88^\circ}{12} \\[5ex] \sin \angle MLN = \dfrac{10 * \sin 88^\circ}{12} \\[5ex] \sin \angle MLN = \dfrac{10(0.999390827)}{12} \\[5ex] \sin \angle MLN = 0.8328256892 \\[3ex] \angle MLN = \sin^{-1}(0.8328256892) \\[3ex] \angle MLN = 56.39010829 \\[3ex] \angle MLN \approx 56^\circ \\[3ex] (ii) \\[3ex] \underline{\triangle LNR} \\[3ex] |NR|^2 = |LN|^2 + |LR|^2 - 2 * |LN| * |LR| * \cos \angle RLN...Cosine\;\;Law \\[3ex] |NR|^2 = 12^2 + 18^2 - 2(12)(18) * \cos 25^\circ \\[3ex] |NR|^2 = 144 + 324 - 432 * 0.906307787 \\[3ex] |NR|^2 = 468 - 391.524964 \\[3ex] |NR|^2 = 76.475036 \\[3ex] |NR| = \sqrt{76.475036} \\[3ex] |NR| = 8.745000629 \\[3ex] |NR| \approx 9\;km \\[3ex] (iii) \\[3ex] Bearing\;\;of\;\;R\;\;from\;\;L \\[3ex] = 50 + \angle MLN + 25...diagram \\[3ex] = 50 + 56.39010829 + 25 \\[3ex] = 131.3901083 \\[3ex] \approx 131^\circ$
(42.)

(43.)

(44.)

(45.)

(46.)

(47.)

(48.)

(49.)

(50.)

(51.)

(52.)

(53.)

(54.)

(55.)

(56.)

(57.)

(58.)

(59.) NZQA The bar-tailed godwit (kuaka) flies long distances.
The diagram below shows how one particular godwit has been tracked from point A to point B and then on to point C.
The godwit flies from A to B, a distance of 1500 km, on a bearing of 128°.
It then turns through 90°, flying a further 800 km to reach point C.
Angle ABC = 90°

(i) Show that the size, u, of angle ABG is 52°.

(ii) Find the bearing of A from C.

$90 + \theta = 128^\circ ...diagram \\[3ex] \theta = 128 - 90 \\[3ex] \theta = 38^\circ \\[3ex] \tau = \theta = 38^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] (i) \\[3ex] u + \tau = 90^\circ ...diagram \\[3ex] u + 38 = 90 \\[3ex] u = 90 - 38 \\[3ex] u = 52^\circ \\[3ex]$

$\tan \phi = \dfrac{|BC|}{|AB|} ...SOHCAHTOA \\[5ex] \tan \phi = \dfrac{800}{1500} \\[5ex] \tan \phi = 0.5333333333 \\[3ex] \phi = \tan^{-1}(0.5333333333) \\[3ex] \phi = 28.07248694^\circ \\[3ex] \theta + \phi = 38 + 28.07248694 = 66.07248694^\circ \\[3ex] (ii) \\[3ex] Bearing\;\;of\;\;A\;\;from\;\;C\;(indicated\;\;by\;\;purple\;\;color) \\[3ex] = 90 + 90 + 90 + \theta + \phi \\[3ex] = 270 + 66.07248694 \\[3ex] = 336.0724869 \\[3ex] \approx 336^\circ$
(60.)

(61.) ACT A highway engineer is using a road map to lay out a detour for the westbound lane of a section of highway that, on the map, is a straight line going east and west.
On the map, the detour goes 4 miles straight north, 1 mile straight west, 2 miles straight north, 6 miles straight west, 3 miles straight south, 1 mile straight east, and finally 3 miles straight south, back to the highway.
According to the map, how many more miles will a westbound driver travel by taking the detour than he would if he could stay on the highway?

$F.\;\; 20 \\[3ex] G.\;\; 14 \\[3ex] H.\;\; 13 \\[3ex] J.\;\; 12 \\[3ex] K.\;\; 6 \\[3ex]$

Let us represent the information on a diagram

$Total\;\;distance\;\;using\;\;the\;\;detour \\[3ex] = 4 + 1 + 2 + 6 + 3 + 1 + 3 = 20\;miles \\[3ex] Distance\;\;covered\;\;directly:\;\;highway\;\;to\;\;highway = 1 + 6 - 1 = 6\;miles \\[3ex] Difference = 20 - 6 = 14\;miles \\[3ex]$ The westbound driver will travel 14 more miles by taking the detour than he would if he could stay on the highway.
(62.)

(63.)

(64.)

(65.)

(66.)