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.​

Answer 1

In ∆AED,

Perpendicular (P)=AE

Base (B)=DE

We know,

$\rm \boxed{tan \theta = \frac{p}{b} }$

Here,

$\rm tan 60 = \frac{AE}{DE} \\ \rm \frac{1}{ \sqrt{3} } = \frac{h - 7}{DE} \\ \rm \boxed{DE = \sqrt{3} (h - 7)}....1$

In ∆DEB,

Perpendicular (P)=EB

Base (B)=DE

We know,

$\rm \boxed{ tan \theta = \frac{p}{b} }$

Here,

$\rm tan45 = \frac{EB}{DE} \\ \rm \: 1 = \frac{7}{DE} \\ \rm \implies \boxed{DE = 7}......2$

From 1 & 2,we get

$\rm\sqrt{3} (h - 7) = 7 \\ \rm \: \sqrt{3} h = 7 + 7 \sqrt{3} \\ \rm \: \implies \boxed{ h = \frac{7(1 + \sqrt{3} }{3}}$

Therefore height of the tower is 7(1+√3)/√3

Answer 2

Step-by-step explanation:

Diagram:- Refer the attachment

Given:-

• The height of the building = 7m

• The angle of elevation of a cable tower from the top of the building is 60°

• The angle of depression of its foot is 45°

To Find:-

• The height of the tower.

Solution:-

Let AB be the height of the building = 7m

And EC be the height of the building.

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

i.e ∠DAE = 60° and ∠CAD = 45°

And, AB = CD = 7m

$\sf \underline{In \: \triangle ABC:-}$

$\sf \implies tan45^{ \circ} = \dfrac{AB}{BC}$

$\sf \implies 1 = \dfrac{7}{BC}$

$\sf \implies BC = 7m$

$\sf \implies AD = BC = 7m$

$\sf\underline{ In \: \triangle ADE:-}$

$\sf \implies tan60^{ \circ} = \dfrac{DE}{AD}$

$\sf \implies \sqrt{3} = \dfrac{DE}{7}$

$\sf \implies DE = 7\sqrt{3}$

$\sf Height \: of \: the\: tower = DE + CD$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 7\sqrt{3} + 7$

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

$\underline{\boxed{\bf \therefore \: Height \: the \: tower = 7(\sqrt{3} + 1)m}}$