# Ex.9.1 Q12 Some Applications of Trigonometry Solution - NCERT Maths Class 10

## Question

From the top of a \(7 \,\rm{m}\) 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.

## Text Solution

**What is Known?**

(i) Height of building \(=7 \,\rm{m}\)

(ii) Angle of elevation of the top of a cable tower from building top \(=60^\circ\)

(iii) Angle of depression of the foot of the cable tower from building top \(= 45^\circ\)

**What is ****Unknown?**

Height of the tower

**Reasoning:**

Let the height of the tower is \(CE\) and the height of the building is \(AB\). The angle of elevation of the top \(E\) of the tower from the top \(A\) of the building is \(60^\circ\) and the angle of depression of the bottom \(C\) of the tower from the top \(A \)of the building is \(45^\circ\). Trigonometric ratio involving building height, tower height, angles and distances between them is \(\tan \theta \)

**Steps:**

Draw \(AD||BC\).

Then,\(\angle DAC=\angle ACB={{45}^{0}}\)(alternate interior angles.)

In \(\Delta ABC\),

\[\begin{align}\tan {{45}^{0}}&=\frac{AB}{BC} \\ 1&=\frac{7}{BC} \\ BC&=7 \end{align}\]

\(ABCD\) is a rectangle,

Therefore, \(BC = AD = 7 \, \rm{m}\) and \(AB = CD = 7 \, \rm{m}\)

In \(\Delta ADE\),

\[\begin{align} { \tan 60 ^ \circ = \frac { E D } { A D } } \\ { \sqrt { 3 } = \frac { E D } { 7 } } \\ { E D = 7 \sqrt { 3 } } \end{align}\]

Height of tower

\[\begin{align}CE&=ED+CD \\ &=7\sqrt{3}+7 \\ &=7(\sqrt{3}+1) \end{align}\]

Height of the tower \(= 7 ( 1 + \sqrt { 3 } ) \,\rm{m}\)