(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 latitu...
Question 1 Report
(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.
Answer Details
(a) Speed along the parallel of latitude
Both points are on latitude \(32^\circ\)N. The difference in longitude is
\[47^\circ + 25^\circ = 72^\circ\]
Distance along a parallel of latitude \(\phi\):
\[d = \frac{\theta}{360}\times 2\pi R\cos\phi\]
\[d = \frac{72}{360}\times 2\times\frac{22}{7}\times 6400\times\cos32^\circ\]
Using \(\cos32^\circ = 0.8480\):
\[d = 0.2\times \frac{44}{7}\times 6400\times 0.8480 \approx 6823\text{ km}\]
Speed \(= \dfrac{\text{distance}}{\text{time}} = \dfrac{6823}{11} \approx 620\) km/h.
Speed \(\approx 620\) km/h.
(b) Height of the kite
Let the foot of the vertical from the kite \(K\) meet the ground at \(O\). \(Q\) is nearer (elevation \(45^\circ\)) and \(P\) is \(15\) m further back (elevation \(35^\circ\)). Let \(OQ = d\) and height \(= h\).
From \(Q\): \(\tan45^\circ = \dfrac{h}{d} = 1 \Rightarrow d = h\).
From \(P\): \(\tan35^\circ = \dfrac{h}{d+15}\).
Substitute \(d = h\):
\[h = (h+15)\tan35^\circ = 0.7002(h+15)\]
\[h - 0.7002h = 10.503 \;\Rightarrow\; 0.2998h = 10.503 \;\Rightarrow\; h \approx 35\text{ m}\]
Height \(\approx 35\) m.
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