From the top of a 7 m 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.

Asked by Pragya Singh | 1 year ago |  85

##### Solution :-

Let AB be the building of height 7 m and EC be the height of the tower.

A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°.

EC = DE + CD

Also, CD = AB = 7 m. and BC = AD

To Find: EC = Height of the tower

In right ΔABC,

tan 45° = $$\dfrac{AB}{BC}$$

1= $$\dfrac{7}{BC}$$

BC = 7

tan 60° =$$\dfrac{DE}{AD}$$

$$\sqrt{3}=$$ $$\dfrac{DE}{7}$$

$$DE=7 \sqrt{3}\;m$$

Now: EC = DE + CD

$$(7\sqrt{3} + 7) = 7(\sqrt{3}+1)$$

Therefore, Height of the tower is $$7(\sqrt{3}+1)\;m$$.

Answered by Abhisek | 1 year ago

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