Q. 194.8( 9 Votes )

From the top of a 7-meter-high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. [Use √3 = 1.732.]

Answer :

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Let AB be the building and CE be the cable tower. We are given that, the angle of elevation of the top of the cable tower from the top of the building is 60° and the angle of depression of the bottom of the tower from the top of the building is 45°. Join C, E to A. Also draw a line from A to EC at the point D parallel to BC. We get two right-angled triangles ABC and AED with right angles at B and D respectively. Also, EAD = 60°. DAC = ACB = 45°. Also given that the height of the building is 7 m. That is AB = 7m. We are to find the height of the tower CE.


We first find the value of BC from ∆ABC using the trigonometric ratio tan.


In ∆ABC,



or,



Now, AD = BC.


In ∆DAE,



or,



So, the height of the cable tower = CE = CD + DE = 7m + 7√3m = 19.12m.


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