(a) Two points X(32°N, 47°W) and Y(32°N, 25°E) are on the earth’s surface. If it takes an aeroplane 11 hours to fly from X to Y along the parallel of latitude, calculate its speed, correct to the nearest kilometre per hour. [Radius of the earth = 6400km; (pi = frac{22}{7})]
(b) Two observers P and Q, 15metres apart observe a kite (K) in the same vertical plane and from the same side of the kite. The angles of elevation of the kite from P and Q are 35° and 45° respectively. Find the height of the kite to the nearest metre.
Explanation
(a)
Difference in longitude = 47° + 25° = 72°
Distance /XY/ = (frac{theta}{360°} times 2pi r)
Where (r = R cos 32°)
(therefore /XY/ = frac{72}{360} times 2 times frac{22}{7} times 6400 times cos 32)
= (2 times frac{22}{7} times 1280 times 0.8480)
= (6822.765km)
(Speed = frac{Distance}{Time})
= (frac{6822.766}{11} = 620.25 km/h)
(b)
( <TQP = 180° – 45° = 135°) (angles on a straight line)
(< QTP = 180° – (135° + 35°) = 10°) (sum of angles in a triangle)
(frac{TQ}{sin 35°} = frac{15}{sin 10°})
(therefore TQ = frac{15 sin 35°}{sin 10°})
= (49.546m)
(therefore sin 45° = frac{TA}{TQ})
(sin 45° = frac{TA}{49.546})
(TA = 49.546 times 0.7071 = 35.03m)
(therefore) The height of the kite = 35m (correct to the nearest whole number)