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Question

The angle of elevation of the top Q of a vertical tower PQ, from a point X on the ground is 60. 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)

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Solution

We have PQ as a vertical tower



In ΔYZQ tan 45=QZYZ

QZYZ=1

QZ=YZ ...(i)

And, in ΔXPQ

tan 60=QPXP

3=QZ+40XP

3=QZ+40YZ ( XP=YZ)

3 QZ=QZ+40 [Using (i)]

3 QZQZ=40

QZ (31)=40

QZ=4031=4031×3+13+1

= 20 (3+1)

= 20 (2.73)

= 54.60 m

QZ=54.6 m

And PQ=XY+QZ=(54.6+40) m=94.6 m.


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