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1995 WAEC Mathematics Theory (a) A man travels from a village X on a bearing of 060° to a...

Mathematics
WAEC 1995

(a) A man travels from a village X on a bearing of 060° to a village Y which is 20km away. From Y, he travels to a village Z, on a bearing of 195°. If Z is directly east of X, calculate, correct to three significant figures, the distance of :

(i) Y from Z ; (ii) Z from X .

(b) An aircraft flies due South from an airfield on latitude 36°N, longitude 138°E to an airfield on latitude 36°S, longitude 138°E. 

(i) Calculate the distance travelled, correct to three significant figures ; (ii) if the speed of the aircraft is 800km per hour, calculate the time taken, correct to the nearest hour.

[Take \(\pi = \frac{22}{7}\), R = 6400km].

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Explanation

(a) < XZY = 180° - (30° + 45°) = 105°

\(\frac{20}{\sin 105°} = \frac{YZ}{\sin 30°}\)

\(YZ = \frac{20 \sin 30°}{\sin 105°} = 10.35km\)

\(\approxeq 10.4km \) (to 3 sig. figs)

(ii) \(\frac{20}{\sin 75} = \frac{XZ}{\sin 45}\)

\(XZ = \frac{20 \sin 45}{\sin 75} = 14.64km\)

\(\approxeq 14.6km\) ( to 3 sig. figs)

(b) Let P and Q be two airfields.

Difference in latitude = 36° + 36° = 72°

(i) Distance between P and Q = \(\frac{72}{360} \times 2 \times \frac{22}{7} \times 6400\)

= \(\frac{22}{7} \times 2560\)

= \(8,045.71km \approxeq 8,050 km\) (3 sig. figs)

(ii) Time taken = \(\frac{8,045.71}{800} = 10.06 hours\)

\(\approxeq 10hours\) (to the nearest hour)


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