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

"Rejoice at all times (in GOD)." - 1 Thessalonians 5:16

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



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.
For ACT Students
The ACT is a timed exam...6060 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
Number 1

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

A.226B.224C.136D.134

Number 1

y=180...2rightsx=44...alternatesBearingofRfromS=y+p=180+44=224
(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.0.1B.0.2C.0.3D.0.5E.0.7

Let us represent this information in a diagram
Number 2

Straightlinedistance=hyphyp2=0.32+0.42...PythagoreanTheoremhyp2=0.09+0.16hyp2=0.25hyp=0.25hyp=0.5miles
(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.195B.135C.225D.045

Number 3

ConsideringPQR|XY|=|YZ|=100m...equidistantYZX=YXZ=p...basesofisoscelesYZX+YXZ+XYZ=180...sumofsofPQRXYZ=90...rightp+p+90=1802p=180902p=90p=902p=45BearingofPfromR...redcolored=180p...sonastraightline=18045=135
(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.225B.090C.270D.135

Number 4

ConsideringPQR|PQ|=|QR|...equidistantQPR=QRP=k...basesofisoscelesQPR+QRP+PQR=180...sumofsofPQRPQR=90...rightk+k+90=1802k=180902k=90k=902k=45k=k1=45...alternatesBearingofPfromR=180+k1=180+45=225
(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)QHS=311185=216
Number 5i

Number 5ii

(ii)LetQS=hh2=5.42+3.522(5.4)(3.5)cos126...CosineLawh2=29.16+12.2537.80.5877852523h2=41.41+22.21828254h2=63.62828254h=63.62828254h=7.976733826QS8.0km
Number 5iii

k+126+185=360...satapointk+311=360k=360311k=49p=p...alternatesp+k=90...rightp=90kp=9049p=41ConsideringQHSsinSQH3.5=sin126h...SineLawsinSQH3.5=0.80901699447.976733826sinSQH=3.50.80901699447.976733826sinSQH=2.831559487.976733826sinSQH=0.3549773055SQH=sin1(0.3549773055)SQH=20.79205349(iii)BearingofSfromQ=90+p+SQH=90+41+20.79205349=151.7920535152
(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 and the angle PQR is 56
Number 6

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


(i)56+w=133...alternatesareequalw=13356w=77(ii)BearingofPfromQ=180+w+56=180+133=313(iii)RP2=2102+29022(210)(290)cos56...CosineLawRP2=44100+841001218000.5591929035RP2=12820068109.69564RP2=60090.30436RP=60090.30436RP=245.133238RP245km
(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.cos1(4020)G.sin1(4020)H.sin1(2040)J.tan1(4020)K.tan1(2040)

Let the degree measure of the angle of the triangle at the vertex where Olivia is standing be θ
Let us represent the information using a diagram
Number 7

SOH:CAH:TOAtanθ=4020θ=tan1(4020)
(8.) CSEC A ship travels from Akron (A) on a bearing of 030 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 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 AˆBC
(iv.) Determine the bearing of Comptin (C) from Bellville (B)


(i.)
Number 8i

(ii.)ABC_BˆAC+30=90...perpendicularangleBˆAC=9030=60|BC|2=|AB|2+|AC|22(|AB|)(|AC|)cosBˆAC|BC|2=902+31022(90)(310)cos60|BC|2=8100+96100558000.5|BC|2=10420027900|BC|2=76300|BC|=76300|BC|=276.2245463|BC|276km(iii.)ABC_sinAˆBC310=sinBˆAC|BC|...SineLawsinAˆBC310=sin60276.2245463sinAˆBC=310sin60276.2245463sinAˆBC=3100.8660254038276.2245463sinAˆBC=268.4678752276.2245463sinAˆBC=0.9719189651AˆBC=sin1(0.9719189651)AˆBC=76.38976123AˆBC76(iv.)
Number 8iv

BearingofComptinfromBellville=θϕ=30...alternatesareequalϕ+τ=AˆBC...diagram30+τ=76.38976123τ=76.3897612330τ=46.38976123θ+τ=180...sonastraightlineθ=180τθ=18046.38976123θ=133.6102388θ133
(9.) WASSCE A town J is 20km from a lorry station, K on a bearing 065
Another town, T is 8km from K on a bearing 155
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.)DistanceofTfromJ=kTKJ=25+65=90k2=j2+t22jtcosTKJ...CosineLawk2=82+2022(8)(20)cos90k2=64+4003200k2=4640k2=464k=464k=21.54065923k22km(ii.)sinθj=sin90k...SineLawsinθ8=sin9021.54065923sinθ=8sin9021.54065923sinθ=8121.54065923sinθ=0.3713906763θ=sin10.3713906763θ=21.80140948θ+ϕ=65...alternatesareequalϕ=65θϕ=6521.80140948ϕ=43.19859052BearingofTfromJ=180+ϕ=180+43.19859052=223.1985905223
(10.) curriculum.gov.mt What is the bearing of Q from P?

Number 10


Number 10

k=k...alternatesareequalk+180+90=325...diagramk+270=325k=325270k=55BearingofQfromP=90+k=90+55=145
(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
The boat then changes course and travels for a distance of 8 km to point C on a bearing of 060

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

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


(a.)
Number 11a

(b.)θ=45...alternatesareequalθ+ϕ=90...complementarysϕ=90θϕ=9045ϕ=45ABC=B=ϕ+60...diagramB=45+60B=105(i.)DistanceAC=bb2=a2+c22accosB...CosineLawb2=82+1522(8)(15)cos105b2=64+2252400.2588190451b2=289+62.11657082b2=351.1165708b=351.1165708b=18.73810478b19km(ii.)BCA=CsinCc=sinBbsinC15=sin10518.73810478sinC=15sin10518.73810478sinC=15(0.9659258263)18.73810478sinC=14.4888873918.73810478sinC=0.773231208C=sin1(0.773231208)C=50.64494187C51
Number 11b

(iii)τ=60...alternatesareequalBearingofAfromC=180+τ+BCA=180+60+50.64494187=290.6449419291
(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.1.4B.2.8C.3.2D.7.7E.8.7

The diagram representing the information is shown:

Number 12

e2=602+402...PythagoreanTheoreme2=3600+1600e2=5200e=5200e=72.11102551mphBut:time=distancespeeddistance=200milesspeed=72.11102551mphtime=20072.11102551time=2.773500981time2.8seconds
(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 PˆQR=90
(b) Calculate QˆPR
(c) Calculate the bearing of R from P


Indicate the bearings on the triangle
Number 13a

(a)BearingofPfromQ=243BearingofRfromQ=153PˆQR=243153...diagramPˆQR=90(b)sinQˆPRQR=sinPˆQRPR...SineLawsinQˆPR54=sin9067sinQˆPR=54sin9067sinQˆPR=54(1)67sinQˆPR=5467sinQˆPR=0.8059701493QˆPR=sin1(0.8059701493)QˆPR=53.70405214
Number 13b

(c)k+243=270...diagramk=270243k=27k+x=90...diagram27+x=90x=9027x=63x=x...alternatesareequalBearingofRfromP=x+QˆPR=63+53.70405214=116.7040521
(14.) WASSCE A bearing of 320° expressed as a compass bearing is

A.N50WB.N40WC.N50ED.N40E

Number 14

320+ϕ=360...satapointϕ=360320ϕ=40CompassBearing=N40W
(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.


DirectDistanceofSfromF_SHF+78=180...sonastraightlineSHF=18078SHF=102SHF_|SF|2=|SH|2+|HF|22|SH||HF|cosSHF...CosineLaw|SF|2=542+14022(54)(140)cos102|SF|2=2916+19600151200.2079116908|SF|2=22516+3143.624765|SF|2=25659.62477|SF|=25659.62477|SF|=160.186219cm|SF|160cm
Number 15

sinHSF|HF|=sinSHF|SF|...SineLawsinHSF140=sin102160.186219sinHSF=140sin102160.186219sinHSF=140(0.9781476007)160.186219sinHSF=136.9406641160.186219sinHSF=0.8548841777HSF=sin1(0.8548841777)HSF=58.74694159LetHSF=kk=k=58.74694159...alternatesareequalDirectBearingofSfromF_Bearing=180+k=180+58.74694159=238.7469416
(16.) JAMB The bearing of a bird on a tree from a hunter on the ground is N72E
What is the bearing of the hunter from the bird?

A.S18WB.S72WC.S72ED.S27EE.S27W

Number 16

θ=72...alternatesareequalBearingofthehunterfromthebird=180+θ=180+72=252ORCompassBearingofthehunterfromthebird=SθW=S72W
(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)GFH=90+40=130|GH|2=|GF|2+|FH|22|GF||FH|cosGFH...CosineLaw|GH|2=502+6522(50)(65)cos130|GH|2=2500+422565000.6427876097|GH|2=6725+4178.119463|GH|2=10903.11946|GH|=10903.11946|GH|=104.4180035|GH|104km(iii)sinHGF|FH|=sinGFH|GH|...SineLawsinϕ65=sin130104.4180035sinϕ=65sin130104.4180035sinϕ=650.7660444431104.4180035sinϕ=49.7928888104.4180035sinϕ=0.4768611459ϕ=sin1(0.4768611459)ϕ=28.48059838θ+ϕ=90...rightθ=90ϕθ=9028.48059838θ=61.51940162BearingofHfromG=θ62
(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.8B.19C.23D.26E.31

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

c2=62+82...PythagoreanTheoremc2=36+64c2=100c=100c=10yardsAlso:d2=52+122...PythagoreanTheoremd2=25+144d2=169d=169d=13yardsDistancewalkedbyLoto_=8+6+12+5=31yardsDistancethatLotocouldhavewalkeddirectly_=c+d=10+13=23yardsHowmanyyardsless_=Difference=3123=8yards
(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.030B.045C.060D.090

Let us represent the information on a diagram

Number 20

tanθ=oppadj...SOHCAHTOAtanθ=60360tanθ=3θ=tan1(3)θ=60




Top




(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

GSM_GSM=294(180+54)GSM=294234GSM=60|MG|2=|MS|2+|GS|22|MS||GS|cosGSM...CosineLaw|MG|2=6352+44822(635)(448)cos60|MG|2=403225+20070456896012|MG|2=603929284480|MG|2=319449|MG|=319449|MG|=565.1981953kmsinMGS|MS|=sinGSM|MG|...SineLawsinMGS635=sin60565.1981953sinMGS=635sin60565.1981953sinMGS=6350.8660254038565.1981953sinMGS=549.9261314565.1981953sinMGS=0.972979277MGS=sin1(0.972979277)MGS=76.65037991BearingofMfromG=360(MGS54)=360(76.6503799154)=36022.65037991=337.3496201
(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)|QR|2=|QH|2+|RH|22|QH||RH|cosQHR...CosineLaw|QR|2=1752+24222(175)(242)cos82|QR|2=30625+58564847000.139173101|QR|2=8918911787.96165|QR|2=77401.03835|QR|=77401.03835|QR|=278.210421|QR|278km
Number 23iii

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

69+82+θ=180...sonastraightline151+θ=180θ=180151θ=29HPR_cosθ=|PH||RH|...SOHCAHTOAcos29=|PH|242242cos29=|PH||PH|=242cos29|PH|=242(0.8746197071)|PH|=211.6579691|PH|212km
(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.) 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.2634B.24326C.11634D.6326E.240

Let us represent the information on a diagram

Number 25

tanθ=oppadj...SOHCAHTOAtanθ=6030tanθ=2θ=tan1(2)...AcuteBearingof;HfromP=180+θ=180+tan1(2)...Reflex 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 tan1(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.

Number 27

(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:

Number 27i

(i)RLT+65=148...bearingofTfromLRLT=14865RLT=83(ii)|RT|2=|LT|2+|LR|22|LT||LR|cosRLT...CosineLaw|RT|2=1802+25022(180)(250)cos83|RT|2=32400+62500(900000.1218693434)|RT|2=9490010968.24091|RT|2=83931.75909|RT|=83931.75909|RT|=289.7097843|RT|290km
Number 27iii

sinLRT|LT|=sinRLT|RT|...SineLawsinθ180=sin83289.7097843sinθ=180sin83289.7097843sinθ=178.6583073289.7097843sinθ=0.6166802676θ=sin1(0.6166802676)θ=38.07411323ψ+θ=65...alternatesareequalψ=65θψ=6538.07411323ψ=26.92588677(iii)BearingofTfromR=180+ψ=180+26.92588677=206.9258868
(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:

Number 29

(b.)(i.)sinZXY|ZY|=sinYZX|YX|...SineLawsinθ11.5=sin10019.5sinθ=11.5sin10019.5sinθ=11.50.98480775319.5sinθ=0.5807840595θ=sin1(0.5807840595)θ=35.50570812θ36ZYX+YZX+ZXY=180...sumofsinaZYX+100+35.50570812=180ZYX=18010035.50570812ZYX=44.49429188(ii.)|XZ|sinZYX=|YX|sinYZX...SineLaw|XZ|sin44.49429188=19.5sin100|XZ|=19.5sin44.49429188sin100|XZ|=19.50.70083820290.984807753|XZ|=13.66634496|XZ|14km
(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.

Number 33

(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)

Number 33i

ϕ=42...alternatesareequalBearingofBfromF(bluecolor)=180+ϕ=180+42=222(ii)|BG|2=|BF|2+|FG|22|BF||FG|cosBFG...CosineLaw|BG|2=322+5522(32)(55)cos103|BG|2=1024+30253520(0.2249510543)|BG|2=4049+791.8277113|BG|2=4840.827711|BG|=4840.827711|BG|=69.57605703|BG|70km
Number 33iii

sinFBG|FG|=sinBFG|BG|...SineLawsinFBG55=sin10369.57605703sinFBG=55sin10369.57605703sinFBG=55(0.9743700648)69.57605703sinFBG=0.7702413136FBG=sin1(0.7702413136)FBG=50.37556355BearingofGfromB(redcolor)=42+FBG=42+50.37556355=92.3755635592
(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.

Number 37

(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:

Number 37i

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

Number 37ii

ϕ=44...alternatesareequal105+θ=180...sonastraightlineθ=180105θ=75(ii)ABC=ϕ+θ...diagramABC=44+75ABC=119(iii)|AC|2=|AB|2+|BC|22|AB||BC|cosABC...CosineLaw|AC|2=522+7222(52)(72)(cos119)|AC|2=2704+51847488(0.4848096202)|AC|2=7888+3630.254436|AC|2=11518.25444|AC|=11518.25444|AC|=107.3231309|AC|107km(iv)sinACB|AB|=sinABC|AC|...SineLawsinACB52=sin119107.3231309sinACB=52sin119107.3231309sinACB=52(0.8746197071)107.3231309sinACB=0.423769083ACB=sin1(0.423769083)ACB=25.07277523β=θ=75...alternatesareequalBearingofAfromC...indicatedbypurplecolor=360(β+ACB)...satapoint=360(75+25.07277523)=360100.0727752=259.9272248260
(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.

Number 39

Find the distance AND bearing of M from L.


DistanceofMfromL=|LM|_|LM|2=|KM|2+|KL|22|KM||KL|cosLKM...CosineLaw|LM|2=(4.4)2+(9.8)22(4.4)(9.8)cos111|LM|2=19.36+96.0486.240.3583679495|LM|2=115.4+30.90565197|LM|2=146.305652|LM|=146.305652|LM|=12.096568733|LM|12.1km To find the bearing of M from L, let us review the diagram

Number 39

sinKLM|KM|=sinLKM|LM|...SineLawsinϕ4.4=sin11112.096568733sinϕ=4.4sin11112.096568733sinϕ=4.4(0.9335804265)12.096568733sinϕ=0.3395800881ϕ=sin1(0.3395800881)ϕ=19.85129284θ+90=126...diagramθ=12690θ=36θ(1st)=θ(2nd)...redcolor=36...alternatesareequalβ+ϕ=θ...diagramβ+19.85129284=36β=3619.85129284β=16.14870716BearingofMfromL=ψ_ψ=270+β...diagramψ=270+16.14870716ψ=286.1487072ψ286
(40.)





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(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°

Number 41

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


(i)LMN_sinMLN|MN|=sinLMN|LN|...SineLawsinMLN10=sin8812sinMLN=10sin8812sinMLN=10(0.999390827)12sinMLN=0.8328256892MLN=sin1(0.8328256892)MLN=56.39010829MLN56(ii)LNR_|NR|2=|LN|2+|LR|22|LN||LR|cosRLN...CosineLaw|NR|2=122+1822(12)(18)cos25|NR|2=144+3244320.906307787|NR|2=468391.524964|NR|2=76.475036|NR|=76.475036|NR|=8.745000629|NR|9km(iii)BearingofRfromL=50+MLN+25...diagram=50+56.39010829+25=131.3901083131
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(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°

Number 59

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

(ii) Find the bearing of A from C.
Show your working clearly.


Number 59i

90+θ=128...diagramθ=12890θ=38τ=θ=38...alternatesareequal(i)u+τ=90...diagramu+38=90u=9038u=52
Number 59ii

tanϕ=|BC||AB|...SOHCAHTOAtanϕ=8001500tanϕ=0.5333333333ϕ=tan1(0.5333333333)ϕ=28.07248694θ+ϕ=38+28.07248694=66.07248694(ii)BearingofAfromC(indicatedbypurplecolor)=90+90+90+θ+ϕ=270+66.07248694=336.0724869336
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(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.20G.14H.13J.12K.6

Let us represent the information on a diagram

Number 61

Totaldistanceusingthedetour=4+1+2+6+3+1+3=20milesDistancecovereddirectly:highwaytohighway=1+61=6milesDifference=206=14miles The westbound driver will travel 14 more miles by taking the detour than he would if he could stay on the highway.
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