Ex.9.1 Q7 Some Applications of Trigonometry Solution - NCERT Maths Class 10

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Question

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a \(20\,\rm{m}\) high building are \(45^\circ\) and \(60^\circ\) respectively. Find the height of the tower.

 

Text Solution

  

What is Known?

(i) Angle of elevation from ground to bottom of the tower \(= 45^\circ\)

(ii) Angle of elevation from ground to top of the tower \(= 60^\circ\)

(iii) Height of the building \(= 20 \,\rm{m}\)

(iv) Tower is fixed at the top of the \(20\,\rm{m}\) high building

 

What is Unknown?

Height of the tower \(\)

Reasoning:

Let the height of the building is \(BC\), height of the transmission tower which is fixed at the top of the building is \(AB\). \(D\) is the point on the ground from where the angles of elevation of the bottom \(B\) and the top \(A\) of the transmission tower \(AB\) are \(45^\circ\) and \(60^\circ\) respectively.

The distance of the point of observation \(D\) from the base of the building \(C\) is \(CD\)

Combined height of the building and tower \(= AC = AB + BC\)

Trigonometric ratio involving sides \(AC,\, BC, \,CD\) and \(∠D\, (45^\circ \rm{and}\, 60^\circ)\) is \(\tan \theta \)

Steps:

In \(\Delta {BCD}\)

\[\begin{align}\tan 45^{\circ} &=\frac{B C}{C D} \\ 1 &=\frac{20}{C D} \\ C D &=20 \mathrm{m} \end{align}\]

In  \(\Delta {ACD}\) 

\[\begin{align} \tan 60^{\circ} &=\frac{A C}{C D} \\ \sqrt{3} &=\frac{A C}{20} \\ A C &=20 \sqrt{3} \mathrm{m} \end{align}\]

Height of the tower, \(AB = AC - BC\)

\[\begin{align}AB&=20\sqrt{3}\text m-20\text m \\ & =20\left( \sqrt{3}-1 \right)\rm m \end{align}\]