(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 ...
Question 1 Report
(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].
Answer Details
(a) Take X as origin with (East, North) components; a bearing \(\theta\) gives direction \((\sin\theta,\cos\theta)\).
\(Y = 20(\sin060°,\cos060°) = (17.32,\ 10.00)\)
From Y on bearing \(195°\), distance \(YZ = d\): \(Z = Y + d(\sin195°,\cos195°) = (17.32 - 0.2588d,\ 10.00 - 0.9659d)\).
Z is due East of X, so its North component is \(0\):
\(10.00 - 0.9659d = 0 \ \Rightarrow\ d = \dfrac{10.00}{0.9659} = 10.35\)
(i) \(|YZ| \approx \mathbf{10.4\text{ km}}\).
East component of Z: \(17.32 - 0.2588(10.35) = 17.32 - 2.68 = 14.64\).
(ii) \(|ZX| \approx \mathbf{14.6\text{ km}}\).
(b) The route is along the same meridian (138°E) from 36°N to 36°S, an angular change of \(36° + 36° = 72°\).
(i) \(\text{Distance} = \dfrac{72}{360}\times 2\pi R = \dfrac{72}{360}\times 2\times\dfrac{22}{7}\times 6400\)
\(= 0.2 \times 40228.57 = 8045.7 \approx \mathbf{8050\text{ km}}\) (3 s.f.).
(ii) \(\text{Time} = \dfrac{8045.7}{800} = 10.06 \approx \mathbf{10\text{ hours}}\).
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