Q26 of 47 Page 1

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 600. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX. (Use √3 = 1.73)


The diagram represents the above problem, with PQ as tower and X any Y as points.


In △XQP





Also, XY = 40 m implies RP = 40 m


QR + RP = QP


QR = QP - RP = (QP - 40) m


In △QYR




IMPLIES QR = YR


QP - 40 = XP [FROM 2 AND YR = XP]






XP = 20(1.73 + 1) = 54.6 m


i.e. distance of tower from the point X is 54.6 m


and


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