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  • The angle of elevation of the top Q of aa vertical tower PQ from a point X on the ground 60. From a point Y , 40m vertically above X, the angle of elevation of the top Q of the tower is 45. Find the height of the tower PQ and the distance PX

The angle of elevation of the top Q of aa vertical tower PQ from a point X on the ground 60. From a point Y , 40m vertically above X, the angle of elevation of the top Q of the tower is 45. Find the height of the tower PQ and the distance PX

 

 

 

 

 
 
 
 
 

Answers (1)

XY = 40 m 

XY = OP

in , \Delta QPX \\\\ \tan 60 \degree = PQ /PX\\\\ \sqrt 3 = PQ / PX\\\\ PQ = \sqrt 3 PX---- (1) \\\\ now , In \: \: \Delta QYD \\\\ \tan 45 \degree = \frac{QO}{YO}\\\\ 1 = QO / YO \\\\ YO = QO

Now \\\\ YO = PX \\\\ PX = QO \\\\ PX = h ---- (2)

Put value of PX in eq (1) 

PQ = \sqrt 3 h \\\\ PQ = PO + QO = \sqrt 3 h \\\\ 40 + h = \sqrt 3 h \\\\ 40 = \sqrt 3 h-h\\\\ h = 40 / \sqrt 3 -1 = 40 /(1.732-1)= 40 / 0.732\\\\ h = 54.64 m \\\\

Height of tower = PQ = PO + h = 40 + 54.64 = 94.64 m 

PX = QO = h 

PX = 54.64 meter

 

Posted by

Ravindra Pindel

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