From the top of a $ 7 \mathrm{~m} $ high building, the angle of elevation of the top of a cable tower is $ 60^{\circ} $ and the angle of depression of its foot is $ 45^{\circ} . $ Determine the height of the tower.

Given:

From the top of a \( 7 \mathrm{~m} \) high building, the angle of elevation of the top of a cable tower is \( 60^{\circ} \) and the angle of depression of its foot is \( 45^{\circ} . \)

To do:

We have to determine the height of the tower.

Solution:  

Let $AB$ be the height of the building and $CD$ be the height of the cable tower.

Let point $A$ be the point of observation.

From the figure,

$\mathrm{AB}=\mathrm{DE}=7 \mathrm{~m}, \angle \mathrm{ADB}=\angle \mathrm{EAD}=45^{\circ}, \angle \mathrm{CAE}=60^{\circ}$

Let the height of the cable tower be $\mathrm{CD}=h \mathrm{~m}$ and the distance between the building and the tower be $\mathrm{BD}=\mathrm{AE}=x \mathrm{~m}$.

This implies,

$\mathrm{CE}=h-7 \mathrm{~m}$

We know that,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { AB }}{BD}$

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

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

$\Rightarrow x=7 \mathrm{~m}$.........(i)

Similarly,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { CE }}{AE}$

$\Rightarrow \tan 60^{\circ}=\frac{h-7}{x}$

$\Rightarrow \sqrt3=\frac{h-7}{7}$              [From (i)]

$\Rightarrow 7\sqrt3=h-7 \mathrm{~m}$

$\Rightarrow h=7\sqrt3+7 \mathrm{~m}$            

$\Rightarrow h=7(\sqrt3+1) \mathrm{~m}$

Therefore, the height of the tower is $7(\sqrt3+1) \mathrm{~m}$.

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Updated on: 10-Oct-2022

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