Q. 244.0( 7 Votes )
The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation is 45°. Find the height of tower PQ. [Take √3 = 1.73.]
PQ is the tower and from a point X on the ground, the angle of elevation of the top of the tower is 60°. Join A and P, and Q and X. Then we get a right-angled triangle QPX with right angle at P and ∠QXP = 60°. Y is a point vertically above X. Then YX = 40 m. Draw a line YZ from Y onto the line QP parallel to PX. We are also given that the angle of elevation of the top of the tower from the point Y is 45°. Join Y and Q. We get a right-angled triangle QYZ with right angle at Z and ∠QYZ = 45°. We are to find the height of the tower PQ.
Clearly, ZY = XP, ZP = XY. Let QZ = x
QZ = ZY = x = 54.79 m.
The height of the tower PQ = PZ + ZQ = 54.79m + 40m = 94.79m.
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