Question

From the top 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.

Answer 1

The height of tower is 7 ( 1 + √3)   meters  

Step-by-step explanation:

Given as :

The height of building = H = 7 meters

The height of cable = h meters

The angle of elevation of the top of a cable tower = 60°

The angle of depression of its foot =  45°

The distance between tower and cable = x meters

According to question

From figure

In Δ AEB

Tan angle = $\dfrac{perpendicular}{base}$

Tan  60° = $\dfrac{AB}{BE}$

√3  = $\dfrac{h-7}{x}$

Or,  x = $\dfrac{h-7}{\sqrt{3} }$                  ...........1

Again

In Δ ECD

Tan angle = $\dfrac{perpendicular}{base}$

Tan  45° = $\dfrac{ED}{CD}$

1 = $\dfrac{7}{x}$

Or,  x = 7                         ..............2

Solving eq 1 and eq 2

$\dfrac{h-7}{\sqrt{3} }$   =  7

Or, h - 7 = 7√3

∴    h = 7 + 7√3

Or,  h = 7 ( 1 + √3)   meters

So, Height = h = 7 ( 1 + √3)   meters

Hence, The height of tower is 7 ( 1 + √3)   meters    Answer

Answer 2

$\huge\fbox{\color{blue}{\colorbox{yellow}{\tt{Answer ♥︎}}}}$

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

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

$\sf{EC = DE + CD}$

$\sf{Also, CD = AB = 7 m \: and \: BC = AD}$

$\sf\bold{\underline{\underline{According \: to \: question,}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{In \: right \: ∆ ABC,}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\boxed{\sf{\pink{tan45° = \frac{AB}{BC} }}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf\rightarrow{1 = \frac{7}{BC} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\underline{\underline{\sf{\bold{\purple{\rightarrow{BC = 7m = AD}}}}}}}$

Also,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{In \: right \: ∆ ADE,}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\boxed{\sf{\pink{tan 60° = \frac{DE}{AD} }}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf\rightarrow{ \sqrt{3} = \frac{DE}{7} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\underline{\underline{\sf{\bold{\purple{\rightarrow{DE = 7 \sqrt{3} m}}}}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: {\boxed{\sf{\pink{Height \: of \: tower = EC =DE + CD = (7 \sqrt{3} + 7)m}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\underline{\underline{\sf{\bold{\purple{∴Height \: of \: tower = 7( \sqrt{3} + 1)m}}}}}}$

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