Question

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is and the angle of depression of its foot is Determine the height of the tower.​

Answer 1

Answer:

$\huge\mathtt{{\colorbox{silver}{ANSWER~~~↴}}}$

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°

EC = DE + CD

also,

CD = AB = 7 m.

and BC = AD A/q,

In right ΔABC, tan 45°

= AB/BC

⇒ 1= 7/BC

⇒ BC = 7 m = AD also, In right ΔADE, tan 60° = DE/AD

⇒ √3 = DE/7

⇒ DE = 7√3 m Height of the tower = EC = DE + CD

= (7√3 + 7) m = 7(√3+1)

▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃

Answer 2

we have to find height (h) of cable tower

So first of all

DEBC is a rectangle as all of the angles are of 90°

Therefore DC=EB=7m

AE=(h-7)m

In ∆AED,

Perpendicular (p)=AE

Base(b)=DE

We know,

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

Here,

$\tan 60 = \frac{AE}{DE }$

We also know,

$\tan \theta = \sqrt{3}$

$\boxed{DE = \frac{h - 7}{ \sqrt{3} } }$

In ∆DEB,

Perpendicular (p)=EB

Base (b)=DE

We know

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

Here,

$\tan45 = \frac{EB}{DE}$

We also know,

$\boxed{ \tan 45 = 1}$

$\boxed{ {DE} = 7 }$

Now equate both equation

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

Height of cable tower=7(√3+1)m