(a) A man travels from a village X on a bearing of 060^\circ to a village Y which is 20km away. From Y, he travels to a village Z, on a bearing of 195^\circ. 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^\circN, longitude 138^\circE to an airfield on latitude 36^\circS, longitude 138^\circE.
(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].
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a) < XZY = 180^\circ - (30^\circ + 45^\circ) = 105^\circ
\(\frac{20}{\sin 105^\circ} = \frac{YZ}{\sin 30^\circ}\)
\(YZ = \frac{20 \sin 30^\circ}{\sin 105^\circ} = 10.35km\)
\(\approxeq 10.4km \) (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^\circ + 36^\circ = 72^\circ
(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)
Source: Myschool Past Questions Archive (Record #54530)