ai. A man travels from a village X on a bearing of 060º to a village Y, which is 20 km away. From Y, he travels to a village Z, on a bearing of 195º. If Z is directly east of X, calculate, correct to three significant figures, the distance of Y from Z
ii. A man travels from a village X on a bearing of 060º to a village Y, which is 20 km away. From Y, he travels to a village Z, on a bearing of 195º. If Z is directly east of X, calculate, correct to three significant figures, the distance of Z from X
bi. An aircraft flies due south from an airfield on latitude 36ºN, longitude 138ºE to an airfield on latitude 36°S, longitude 138ºE. Calculate the distance travelled, correct to three significant figures.
ii. An aircraft flies due south from an airfield on latitude 36ºN, longitude 138ºE to an airfield on latitude 36ºS, longitude 138ºE. If the speed of the aircraft is 800 km per hour, calculate the time taken, correct to the nearest hour. [Take π = \(\frac{22}{7}\), R = 6400 km]
ai. ∠Y = 270º - (195° + 30°) = 270° - 225° = 45º
∠Z = 180º - (30° + 45º) = 105º
Let the distance of Y from Z be k
\(\frac{\text{k}}{sin30º} = \frac{20}{sin 105º}\)
k sin 105º = 20 sin 30º
k = \(\frac{20 sin 30°}{ sin 105°}\) = \(\frac{20 × 0.5 }{sin 105°}\) = \(\frac{10}{0.9659}\) = 10.35
⇒ k = 10.4 km (3. s.f.)
ii. Let the distance of Z from X be t
Using the sine rule
\(\frac{\text{t}}{sin 30º} = \frac{20}{sin 105º}\)
t sin 105° = 20 sin 45°
t = \(\frac{20 sin 45º}{sin 105º}\)
t = 14.6km to 3. s.f
bi. Latitudinal difference = 36° + 36° = 72°
Distance PQ = Length of arc PQ
= \(\frac{θ}{360}\) × 2πR = \(\frac{72}{360}\) × 2 × \(\frac{22}{7}\) × 6400 km
= \(\frac{88 \times 640}{ 7 }\) = \(\frac{56320}{7}\) = 8045.7 km = 8050 km (3 s.f.)
ii. Speed of aircraft = 800 km/hr
Time taken = \(\frac{\text{Total distance}}{\text{Speed}}\) = \(\frac{8050}{800}\) = 10.05 hrs
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