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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

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Solved Examples on Bearings and Distances

Samuel Dominic Chukwuemeka (SamDom For Peace) Pre-requisites:
(1.) Angles
(2.) Bearings and Distances
(3.) Triangles

Verify your answers with these Calculators as applicable.
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For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
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Decimals are used appropriately rather than commas
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Solve all questions.
Draw diagrams as applicable.
Show all work.

(1.) JAMB
Number 1

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] $

Number 1

$ 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
Number 2

$ 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] $

Number 3

$ 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] $

Number 4

$ 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
Number 5
(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] $
Number 5i

Number 5ii

$ (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] $
Number 5iii

$ 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 $56^\circ$
Number 6

(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
Number 7

$ 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.

Number 8

(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.)
Number 8i

$ (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] $
Number 8iv

$ 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
Number 9

$ (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?

Number 10


Number 10

$ 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.)
Number 11a

$ (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] $
Number 11b

$ (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.)

$ $
(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.

Number 13

(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
Number 13a

$ (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] $
Number 13b

$ (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] $

Number 14

$ 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

Number 15

Find the direct distance and bearing of S from F.
Show your working clearly.


$ \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] $
Number 15

$ \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] $

Number 16

$ \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)
Number 17

$ (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.
Number 19

$ 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.)





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(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.

Number 21

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


Let us represent the information in the diagram.
Number 21

$ \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

Number 23

(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)
Number 23i

$ (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] $
Number 23iii

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
Number 24a

Kemal wants to measure the bearing of A from B
He draws two lines and measures the angle between them.
Number 2b

Kemal says that the bearing of A from B is 100°
Is his method correct?
Give a reason for your answer.

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

(c) This map shows an airport, E, on an island.
Number 24c

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
Number 24a


(b)
Number 24b

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
Leave your answer in km
(25.)

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