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2000 WAEC Mathematics Theory (a) Two places X and Y on the equator are on longitudes 67°E and 123°E...

Mathematics
WAEC 2000

(a) Two places X and Y on the equator are on longitudes 67°E and 123°E respectively. (i) What is the distance between them along the equator? (ii) How far from the North pole is X? [Take \(\pi = \frac{22}{7}\) and radius of earth = 6400km].

(b)  In the diagram, PQR is a circle centre O. N is the mid-point of chord PQ. |PQ| = 8cm, |ON| = 3cm and < ONR = 20°. Calculate the size of < ORN to the nearest degree.

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Explanation

(a)   Longitude difference = 123° - 67° = 56°

(i) Distance between X and Y along the equator = \(\frac{56}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 0\)

= \(\frac{56}{360} \times 2 \times \frac{22}{7} \times 6400\)

= \(6257.78km \approxeq 6258km\)

(ii) Latitude difference =  90° 

Distance from the North pole = \(\frac{90}{360} \times 2 \times \frac{22}{7} \times 6400\)

= \(10057.14km \approxeq 10057km\)

(b) 

\(OP^{2} = PN^{2} + ON^{2}\)

\(4^{2} + 3^{2} = 25\)

\(OP = \sqrt{25} = 5cm\)

\(OP = OR = 5cm\)

\(\Delta NOR = scalene\)

\(\frac{OR}{\sin 20} = \frac{3}{\sin < ORN}\)

\(\frac{5}{\sin 20} = \frac{3}{\sin < ORN}\)

\(\sin < ORN = \frac{3 \times \sin 20}{5}\)

\(\sin < ORN = 0.2052\)

\(< ORN = \sin^{-1} (0.2052)\)

\(< ORN = 11.84° \approxeq 12°\)

 


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WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
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