From the top of a 7 m high building,

Question:

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.

Solution:

Let OC be the tower of height m and m high building makes an angle of elevation of top of cable wire is and an angle of depression from the its foot is .

Let $B C=x, A D=x$ and $C D=7, A B=7$ and $\angle O A D=60^{\circ}, \angle A C B=45^{\circ}$

So we use trigonometric ratios.

In a triangle $A B C$,

$\Rightarrow \quad \tan C=\frac{A B}{B C}$

$\Rightarrow \quad \tan 45^{\circ}=\frac{7}{x}$

$\Rightarrow \quad 1=\frac{7}{x}$

$\Rightarrow \quad x=7$

Again in a triangle,

$\Rightarrow \quad \tan A=\frac{O D}{A D}$

$\Rightarrow \quad \tan 60^{\circ}=\frac{h}{x}$

$\Rightarrow \quad \sqrt{3}=\frac{h}{7}$

$\Rightarrow \quad h=7 \sqrt{3}$

$\Rightarrow \quad H=h+7$

$\Rightarrow \quad H=7 \sqrt{3}+7$

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

Hence the height of tower is $7(\sqrt{3}+1) \mathrm{m}$.

 

 

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