Questions
 Town X is 13.5km from town Y on a bearing of 028^{o}. A matatu leaves y at 7:35 a.m towards a bearing of 080^{o}. The matatu is at point Z due south of X at 8:55a.m.
 Calculate the average speed of the matatu from Y to Z
 If the matatu continues on the same bearing, calculate the distance it covers from Z when it is East of X.
 Three towns X, Y and Z are such that Y is 500km on a bearing of 315^{o} from X. Z is on a bearing of 230^{o} from X. given that the distance between Y and Z is 800km.
 using a scale of 1cm to represent 100km, draw a scale diagram to show the position of the Towns
 Find the bearing of;
 X from Z
 Z from Y
 Use the scale drawing to find the distance from X to Z
 Two aeroplanes S and R leave an airport at the same time. S flies on the bearing of 240^{o}
at 750Km/h while R flies due East at 600Km/hr.
 Calculate the distance of each aeroplane after 30minutes
 Using a scale of 1cm to represent 50km make an accurate scale drawing to show the positions of the aeroplanes after 30minutes

 Use the scale drawing to find the distance between the two aeroplanes after 30minutes
 If each aeroplane landed after 30minutes and S received a signal to join R in 45minutes. Find its speed.
 Determine the bearing of:
 S from R
 R from S

 The table below gives a field book showing the results of a survey of a section of a piece of land between A and E. All measurements are in metres.
 Draw a sketch of the land.
 Calculate the area of this piece of land.
 Three towns A B and C are situated such that town A is 40km from B on a bearing of 280^{o}. C is 60km from B on a bearing of 130^{o}. Another town D is only 10km from C on a bearing of 210^{o}.
 Drawing accurately and using a scale of 1cm to 10km find the distance from A to C and the bearing of A from C.

 Distance of B from D
 Distance of A from D
 Bearing of A from D
 Bearing of C from D
 A train left Naivasha for Nakuru at 1000hours. It traveled at an average speed of 45km/h and reached Gilgil after 40minutes. It then covered the remaining 50km in 1½ hours. A second train left Nakuru for Naivasha at 1015 hours and arrived at Gilgil at the same time as the first train arrived at Nakuru.
 Using a scale of 1cm to represent 10minutes in the time axis and 1cm to represent 10km on the distance axis, draw on the same axes the graphs to show the movement of the two trains.
 use your graph to find;
 the distance between Naivasha and Nakuru
 the time at which the train met
 calculate the average speed, in km/h of the second train
 On a certain map, a road 20km long is represented by a line 4cm long. Calculate the area of a rectangular plot represented by dimensions 2.4cm by 1.5cm on this map – leaving your answer in hectares.
 A port B is on a bearings of 080^{o} from a port A and at a distance of 95km. A submarine is stationed at a port D, which is on a bearing of 200^{o} from A, and a distance of 124km from B. A ship leaves B and moves directly southwards to an island P, which is on a bearing of 140^{o }from A. The submarine at D on realizing that the ship was heading for the island P, decides to head straight for the island to intercept the ship.
 Using a scale of 1cm to represent 10km draw a diagram to show the positions of A,B,D, and P
 Hence;
Determine; the distance from A to D
 the bearing of the submarine from the ship when the ship was setting off from B
 the bearing of the island P from D
 the distance the submarine had to cover to reach the island P
 Use a scale of 1cm represents 50km in these questions. Five towns A, B, C, D and E are situated such that A is 200 km from B on a bearing of 050° from E. C is 300 km from B on a bearing of 150° from B. D is 350km on a bearing of 240° from C. E is 200km from D and the bearing of D from E is 100°
 Draw the diagram representing the positions of the towns
 From the diagram, determine;
 The distance in km of A from E
 The bearing of D from B
 Four towns P, Q, R & S are such that P is 280 km North of R, S is 190 km from R on a bearing of 310^{o} and Q is 240 km from P on a bearing of 105^{o}.
 Using scale of 1 cm rep. 50 km, locate the four towns.
 Find;
 distance SQ.
 Bearing of S from Q.
 The shortest distance between P and side QR.
 Four ships are at sea such that a streamliner S is 150km on a bearing of 025° from a cargo ship C. A trawler T is 300km on a bearing of 145° from the cargo ship and a yacht Y is due West of C and on a bearing of 300° from T.
 Using a scale of 1cm= 50km, draw on accurate scale drawing showing the positions of S, C, T and Y
 By measurement from your scale drawing determine:
 The distance and bearing of Y from S
 The distance ST
 The distance YT
 A tea farm in Kakamega forest was surveyed and the results were recorded in the surveyors note book as shown below. The measurements are in meters.
Using a scale of 1:25, draw the map of the plot and hence calculate the area of the plot in Hectares.  The information below shows the entries in a surveyor’s field book after a survey of a farm. XY = 280m is the baseline. All measurements are in metres.
 Use a scale of 1cm represents 20m to draw the map of the farm
 Estimate the area of the farm in hectares
 If the point Y lies due north of X, find correct to 1 decimal place, the :
 Bearing of E from X
 Distance of E from X
 The measurements of a flower garden were recorded in a surveyor’s field book as shown.
Draw a sketch of the field and find its area. (Measurements are in m)  A map has a scale 1:40,000:
 Calculate the distance between two points on the ground if the corresponding distance shown on the map is 3.25cm
 Calculate the area in the map of woodland which occupies 36ha on the ground
 Three scouts John, Peter and Samwel stand on three adjacent peaks of equal altitude on mountain range. The distance between John and Peter is 800metres and the bearing of Peter from John is 020^{o}. The distance between John and Samwel is 1500metres, and the bearing of Samwel from John is 320^{o}.
 Calculate the bearing of John from Peter
 Calculate:
 the distance
 the bearing of Samwel from Peter
 The figure below represents a surveyor’s sketch of a plot of land. Calculate the area of the plot in square metres given that XY = 50m, XK = 20m, XM = 25m, XL = 35m, KA = 40m, MD = 38m and LB = YC = 60m.
 Two boats P and Q are located 30km apart; P being due North of Q. An observer at P spots a ship whose bearing he finds as S 56^{o}E from Q, the bearing of the same ship is 038^{o}. Calculate the distance of the ship from Q to 2 decimal places
 A map is drawn to scale of 1:100,000. What area in km², is represented by a rectangle measuring 4.5cm by 5.4 cm
 Town B is 180 km on a bearing of 050^{o} from town A. Another town C is on a bearing of 110^{o }from town A and on a bearing of 150^{o} from town B. A fourth town D is 240 km on a bearing of 320^{o} from A. Without using a scale drawing, calculate to the nearest kilometer.
 The distance AC
 The distance CD
Answers

 ^{YZ}/_{Sin 28o} = ^{13.5}/_{sin 100o}
Duration of travel = 8:55am  7:35am
= ^{4}/_{3}
Speed = 6.436 ÷ ^{4}/_{3}
= 4.827 km/hr ^{13.5}/_{Sin 10o} = ^{(6.436 + ZQ)}/_{Sin 118o}
Sin 118o
6.436 + ZQ = 13.5 x sin118^{o}= 68.659
ZQ= 68.6596.436
= 62.223
 ^{YZ}/_{Sin 28o} = ^{13.5}/_{sin 100o}



 049 ± 1
 190 ± 1
 6.7 ± 0.1
670 ± 10



 Distance covered by s
= (750 x ½ )km = 375 km
Distance covered by R
= (600 x ½ ) km = 300 km
 Distance covered by s

 Distance between the two aeroplanes
= 12.5 x 50 = 625 ± 5 km  Speed = (^{625}/_{45} x 60) km/hr
833^{1}/_{3} km/h
 Distance between the two aeroplanes

 Bearing of S from R = 225^{o}
 The bearing of R from S = 72^{o}


Area A: ½ x 25 (33 + 21) = 675
Area B: ½ x 40 (21 x 42) = 1260
Area C: ½ x 30 x 42 = 630
Area D: ½ x 25 x 40 = 500
Area E: ½ x 5 (40 + 25) = 162.5
Area F: ½ X 60 (25 + 36) = 1830
Area G: ½ x 5 x 36 = 90 √
= 5,147.5m^{2} 
 A to C = 96 ± 1 km
Bearing = 300^{o} 
 62 ± 1km
 97 ± 1 km
 304^{o}
 030^{o}
 A to C = 96 ± 1 km

 Graph

 80 km
 11.06a.m
 Average speed of the 2^{nd} train
Time taken = 80 ÷ 1^{11}/_{12}
= 80 x ^{12}/_{23}
= 41.74km/h
 L.S.F = ^{4}/_{2000000 }= ^{1}/_{500000}
A.S.F = ^{1}/_{5 x 105} = ^{1}/_{2.5 x 1011}
Area of rectangle = (2.4 x 1.5) cm^{2}
= 3.6cm^{2}
Actual area
=3.6 x 2.5 x 10^{11} ha
100 x 10000
= 9 x 10^{5}
= 900,000ha 
 Δ ABD correctly constructed
Δ ABP 
 AD = 4.5 + 0.1cm
Distance A + D
= 4.5 X 10 = 45km  Bearing of (i) from B
= 241 + 1  Bearing P from D
= 123 = 2  Dp = 12.9 + 0.2 am
Distance D + P = 12.9 X 10
= 129 km
 AD = 4.5 + 0.1cm
 Δ ABD correctly constructed


 6.8 + 0.1cm
Distance Ae = 340 + 5 km  180 + 18 = 198 + 2
 6.8 + 0.1cm



 SP = 7.8 x 50 = 390 km ± 5 km
 S & Q = 255^{o} ± 1^{o}
 4 x 50 = 200 km + 5 km


 Scale = 50km
Drawing accurately<NCE = 25^{o}
<NCT = 145^{o}
<NTY = 90^{o}
Lines drawn //  By measurement:
 Distance SY = 6.9 x 50 = 345 ± 5km
Bearing Y For S = 360^{o} – 114 = 246±1^{o}  distance ST = 7.9 x 50 = 39.5 ± 5km
 distance YT = 9.8 x 50 = 490 ± 5km
 Distance SY = 6.9 x 50 = 345 ± 5km
 Scale = 50km
 Area of A = ½ x 50 x 60 = 1500m^{2}
B = ½ x 70 x 60 = 2100m^{2}
C = ½ (60 +80) x 120=11050m^{2}
D = ½ x 80 x 80 = 3200m^{2}
F = ½ x 10 x 70 = 350m^{2}
Total area = 26600m^{2}
Ha = ^{26600}/_{10000} = 2.66ha 

Total area = area (1) + (2) + (3) + (4) +(5) + (6) + (7)
Area (1)= ½ x 90 x 100 = 4500m^{2}
(2) = (100 + 105)10 = 10250m^{2}
2
(3) = ½ x 90 x 105 = 4725m^{2}
(4) = ½ x 50 x 110 = 2750m^{2}
(5) = ½ x (110 + 45)70 = 5425m^{2}
(6) = (45 + 95) 120 = 8400m^{2}
(7) = ½ x 40 x 95 = 1900m^{2}
Total area = 37,950m^{2}
In hectares = ^{37950}/_{10000} ha = 3.795ha 
 bearing of E from x is 0.25 ± 1^{o}
 Distance Ex = (12.8 x 0.1 x 20m) = 256 ± 2m

Area A = ½ x 170 x 80 = 6800
B = ½ x 80 x 80 = 3200
C = ½ x 10 x 70 = 350
D = ½ x 170 x 130 = 11050
E = ½ x 70 x 60 = 2100
Total = 23,500 m2 
 ^{L.s.f}/_{40,000} = 1
^{1}/_{40,000} = ^{3.25}/_{x}
x = 130,000cm  A.s.f = (^{1}/_{40000})^{2}
(^{1}/_{40000})^{2} = ^{x}/_{36000000}
x = 0.0225cm^{2}
 ^{L.s.f}/_{40,000} = 1

 bearing = 180 + 20 = 200^{o}
 a^{2} = 1500 +
a^{2} = b^{2} + c^{2} – 2bc cos A
a^{2} = 1500^{2} + 800^{2} – 2 x 1500 x 800cos 60
= 2250000 + 640000 – 1200000
= 1690000
a = 1300m  ^{1300}/_{Sin 60} = ^{1500}/_{sin c}
1300 sin c = 1500 sin 60
Sin c = 1500 sin 60
1300
= 0.9993
c = 87.79^{o}
c = 87.80^{o}

A of Δ XYD = ½ x 50 x 38 = 950m^{2}
A of XBCY = ½ (50 + 15) 60
= ½ x 65 x 60
= 1950m^{2}
Total A = (950 + 1950)m^{2}
= 2900m^{2}  B1 for 86^{o}
^{30}/_{Sin 86o} = ^{Q5}/_{Sin 56o}
QS =
30sin 56^{o}
Sin 86^{o}
= 24.93km  1cm for 100000cm
1cm^{2} = (100000cm)^{2}
Area = 5.4 x 4.5 x 100000 cm^{2}
= 5.4 x 4.5 x 100000 x 100000Km^{2}
100000 x 100000
= 24.3km^{2} 
Use sine rule in ΔABC
^{X}/_{Sin 80o} = ^{180}/_{Sin 40o} => x = ^{180 sin 80o}/_{sin 40o}
= 275.8
Hence AC = 276 km Use the cosine rule in Δ AD when < DAC = 150^{o}
y^{2} = 240^{2} + 276^{2} – 2 x 240 x 276 cos 150^{o}
= 576000 + 76180 – 132 480 (cos 30^{o})
= 133776 + 114731 = 248507
y = √248507
= 498.5
Hence CD = 499 km  Using sine rule in ΔABC we have
BC = 180
Sin 60^{o} sin 40^{o} BC =
180 sin 60
Sin 40
= 242.5
= 243 km
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