Ex.9.1 Q8 Some Applications of Trigonometry Solution - NCERT Maths Class 10

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Question

A statue, \(1.6\,\rm{m}\) tall, stands on the top of a pedestal, from a point on the ground. The angle of elevation of the top of the statue is \(60^\circ\) and from the same point the angle of elevation of the top of the pedestal is \(45^\circ\). Find the height of the pedestal.

Text Solution

  

What is Known?

(i) Height of statue \(= 1.6 \,\rm{m}\)

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

(iii) Statue stands on the top of the pedestal.

(iv) Angle of elevation of the top of the pedestal (bottom of the statue) \(= 45^\circ\)

 

What is Unknown?

Height of the pedestal

Reasoning:

Let the height of the pedestal is \(BC\), height of the statue, stands on the top of the pedestal, 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 statue \(AB\) are \(45^\circ\) and \(60^\circ\)respectively.

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

Combined height of the pedestal and statue \(AC = AB + BC\)

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

Steps:

In \(\Delta BCD\),

 \[\begin{align}\tan {45^0} &= \frac{{BC}}{{CD}}\\1 &= \frac{{BC}}{{CD}}\\BC &= CD\qquad \left( {\rm{i}} \right)\end{align}\]

In  \(\Delta {ACD}\)  

\[\begin{align}\tan {\rm{ }}{60^0} &= \frac{{AC}}{{CD}}\\\tan {\rm{ }}{60^0} &= \frac{{AB + BC}}{{CD}}\\\sqrt 3  &= \frac{{1.6 + BC}}{{BC}}\qquad\left[ {{\rm{from}}\left( {\rm{i}} \right)} \right]\\\sqrt 3 BC &= 1.6 + BC\\\sqrt 3 BC - BC &= 1.6\\BC\left( {\sqrt 3  - 1} \right) &= 1.6\\BC &= \frac{{1.6}}{{\sqrt 3  - 1}} \times \frac{{\sqrt 3  + 1}}{{\sqrt 3  + 1}}\\&= \frac{{1.6\left( {\sqrt 3  + 1} \right)}}{{3 - 1}}\\&= \frac{{1.6\left( {\sqrt 3  + 1} \right)}}{2}\\&= 0.8\left( {\sqrt 3  + 1} \right)\end{align}\]

Height of pedestal \(BC\) \(=0.8(\sqrt{3}+1) \mathrm{m}\)