Published on February 23rd 2025 | 6 mins , 1030 words
Latitudes and Longitudes
Form 4 KCSE Paper 2 Question
18. An aeroplane that moves at a constant speed of 600 knots flies from town \( P(14^\circ \text{N}, 30^\circ \text{W}) \) southwards to town \( Q(5^\circ \text{S}, 30^\circ \text{W}) \) taking \( 3\frac{1}{2} \) hours. It then changes direction and flies along the latitude to town \( R \) on longitude \( 60^\circ \text{E} \). Given \( \pi = 3.142 \):
(a) Calculate:
(i) The value of \( x \). (3 marks)
Let the distance between \( P \) and \( Q \) be \( n \) nm.
\(
600 = \frac{n}{3\frac{1}{2}} \quad \Rightarrow \quad n = 600 \times \frac{7}{2} = 2100 \text{ nm}.
\)
Latitude difference:
\(
35^\circ - 14^\circ = 21^\circ.
\)
(ii) The distance between town \( Q \) and town \( R \) along the parallel of latitude in km. (3 marks)
Using the formula:
\(
r = R \cos \theta \quad \Rightarrow \quad 6370 \cos 21^\circ = 5948.42 \text{ km}.
\)
Distance along the parallel:
\(
= \frac{2\pi \times 6371 \times \cos 21^\circ}{360} \times (60^\circ - 30^\circ) = 9471.80 \text{ km}.
\)
(b) \( S \) is an airport situated at \( (5^\circ \text{N}, 120^\circ \text{W}) \):
(i) Calculate the time the aeroplane would take to fly from \( R \) to \( S \) following a great circle through the South Pole. (3 marks)
\(
\text{Total angle difference} = 159^\circ + 5^\circ = 164^\circ.
\)
\(
\text{Distance } D = 60 \times 164 = 9840 \text{ nm}.
\)
\(
\text{Time} = \frac{9840}{600} = 16.4 \text{ hrs}.
\)
(ii) The local time the plane arrived at \( S \) if the local time at \( R \) is 12:20 pm when the plane departed. (2 marks)
\(
12:20 + 16 \text{ hrs } 24 \text{ mins } = 4:44 \text{ am (next day)}.
\)
An airplane leaves town A (83°N, 155°W) to town B (40°N, 25°E) using the shortest route at a speed of 450 knot (Take π = 22/7 and radius of the earth R = 6370km).
(a) (i) Calculate distance between A and B in nautical miles. (2 marks)
Distance (nm) = \(60 \theta \)
\( = 60 \times 57 = 3420 \text{ nm}\)
(ii) Calculate the time taken to travel from town A to B
\(t = \frac{D}{S} \)
\(= \frac{3420}{450}\)
\( = 7.6 \text{ hrs}\)
\( = 7 \text{ h } 36 \text{ min}\)
(b) From B, the plane flies westwards along the latitude to town C (40°N, 13°W).
Calculate the distance BC in kilometers. (3 marks)
Distance(Km) \(= \frac{\alpha}{360} 2 \pi R \cos \text{lat} \)
\(= \frac{38}{360} \times 2 \times \frac{22}{7} \times 6370 \cos 40^\circ \)
\(= 3237.64 \text{ Km}\)
(c) From town C, the plane took off at 3:10 pm towards town D \((10^\circ N, 13^\circ W)\) at the
same speed. At what time did the plane land at D?(3 marks)
Distance(nm) \(= 60 \theta \)
\(= 60 \times 30 \)
\(= 1800 \text{ nm} \)
\(t = \frac{1800}{450} \)
\(= 4 \text{ h}\)
\( \begin{aligned}
& 3.10 \\
& \underline{4.00} \\
& 7.10 \text{ pm}
\end{aligned} \)
KCSE 2022 Maths Paper 2 Queston
15. An aircraft took off from an airport \(A(0^{\circ},40^{\circ}W)\) at 1100 h local time. The aircraft landed at airport \(B(0^{\circ},65^{\circ}W)\) at 1200 h local time.
Determine the speed of the aircraft in knots. (4 marks)
\(S=\frac{D}{\tau}\rightarrow\frac{nm}{h}=Knots\)
\(D=\theta\times60^{\circ}\)
\((65^{\circ}-40^{\circ})=25^{\circ}\times60=1500nm\)
\(1^{\circ}=4mins\)
\(25^{\circ}=100mins \)
\(= 1hr \quad 40mins\)
\(A\rightarrow1100hrs - B\rightarrow0920hrs\)
\(\begin{aligned}
\rightarrow & 1200 \text{ hrs} \\
& \underline{0920} \\
& 240 \\
& 2 \text{ hrs } \quad 40 \text{ Mins}
\end{aligned} \)
\(=2\frac{2}{3} \text{hrs}\)
\(=\frac{8}{3} \text{hours} \)
Speed = \(\frac{1500 nm}{\frac{8}{3} hrs } \)
\(= 563.5 \quad knots\)
A ship left point P(10°S, 40°E) and sailed due East for 90 hours at an average speed of 24 knots to a point R. (Take 1 nautical mile (nm) to be 1.8531 km and radius of the earth to be 6370 km)
(a) Calculate the distance between P and R in:
(i) nm;2 (1 mark)
Distance = Speed x Time
\(24 \, \frac{\text{nm}}{\text{hr}} \times 90 \, \text{hrs} = 2160 \, \text{nm}\)
(ii) km. (1 mark)
If 1 nm = 1.853 km
\(2160 \, \text{nm} = 4002.48 \, \text{km}\)
(b) Determine the position of point R. (5 marks)
Dist = \(60 \theta \cos \alpha\)
2160 = \(60 \theta \cos 10\)
Dist = \(\frac{\theta}{360} 2\pi R \cos \alpha\)
4002.5 = \(\frac{\theta}{360} 2\pi R \cos 10\)
\(\frac{2160}{60 \cos 10} = \theta\)
\(\theta = 4002.5 \times 360 \times \frac{1}{2} \times \frac{1}{\pi} \times \frac{1}{R}\)
\(36.55^{\circ} = \theta\)
\(\theta = 36.54^{\circ}\)
\(R = [10^{\circ}S, 76.55^{\circ}E]\)
(b) KCSE 2007, PP2, Q13
Two places A and B are on the same circle of latitude north of the equator. The longitude of A is 118°W and B is 133°E. The shorter distance between A and B measured along the circle of latitude is 5422 nautical miles. Find, to the nearest degree, the latitude on which A and B lies. (3 marks)
(c) KCSE 2009, PP2, Q13
Points P(40°S, 45°E) and Q(40°S, 60°W) are on the surface of the earth. Calculate the shortest distance along a circle of latitude between the two points. (3 marks)
(a) Calculate the shortest distance in km between M(15°S, 40°W) and N(45°S, 140°E). (3 marks)
(b) KCSE 2014, Maths PP2, Q13
The shortest distance between two points A(40°N, 20°W) and B(θ°S, 20°W) on the surface of the earth is 8 008 km. Given that the radius of the earth is 6 370 km, determine the position of B. (Take π = 22/7). (3 marks)
Solution
Note longitude 40°W = longitude 140°E
Since 40 + 140 = 180°
So M and N are on the same longitude.
θ = 180 – (15 + 45)
= 180 – 60 = 120°
Arc length along longitude = (θ/360) × (2πR)
= (120/360) × 2 × (22/7) × 6 370
= 13 346.67 km
(b) Extract the longitude:
(40 + θ)/360 × 2πRₑ = 8008
40 + θ = (8008 × 7 × 360) / (2 × 22 × 6 370)
θ = 72 – 40
= 32°S
B(32°S, 20°E)
Muranga County Mock 2024 Maths Paper 2
(a) (i) Taking the radius of the earth, R=6370km and 𝜋 = 22, calculate the shortest
7
distance between two cities P(600N, 290W) and Q(600N, 310E) along the parallel of
latitude. (3marks)
(ii) If it is 1200hrs at P, what is the local time at Q (3marks)
(b) An aeroplane flew due south from a point A(600N, 450E) to a point B, the distance covered by the aeroplane was 8000km, determine the position of B. (4marks)