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From a point on the ground, the angle of elevation of the top of a tower is observed to be 60 degree. From a point 40m vertically above the first point of observation, the angle of elevation of the top of the tower is 30 degree. Find the height of the tower and its horizontal distance from the point of observation 

 

 

 

 
 
 
 
 

Answers (1)

Let Point B is above point A 40 meter 

AB = 40 m 

 \angle QAP= 60 \degree \\\\ \angle QBR = 30 \degree

AB = PR = 40 m \\\\ AP = BR \\\\ In \: \: \Delta QAP

 

\tan60 = \frac{PQ}{AP} \\\\ \sqrt 3 = \frac{PR + QR }{AP} \\\\ \sqrt 3 = \frac{40 + QR w}{ AP }\\\\ \sqrt 3 AP = 40 + QR ---- ( 1) \\\\

In \: \: \Delta QBR \\\\ \tan 30 \degree = QR /BR \\\\ 1/\sqrt 3 = QR /AP \\\\ AP= \sqrt 3 QR ---- (2)

putting value of AP in (1) 

\sqrt 3 ( \sqrt 3 QR ) = 40 + QR \\\\ 3QR - QR = 40 \\\\ 2 QR = 40 \\\\ QR = 40/2 \\\\\ QR = 20

Length of the tower (PQ )= PR + QR = 40 + 20 = 60 m 

Distance of tower feet from point A = AP 

AP = \sqrt 3 QR

=\sqrt 3 \times 20 \\\\ =20 \sqrt 3 m

Posted by

Ravindra Pindel

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