Ex.9.1 Q16 Some Applications of Trigonometry Solution - NCERT Maths Class 10

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Question

The angles of elevation of the top of a tower from two points at a distance of \(4\,\rm{m}\)  and \(9\,\rm{m}\)  from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is \(6\,\rm{m.}\)

 

Text Solution

  

What is Known?

Angle of elevation of the top of a tower from two points at a distance of \(4\,\rm{m}\) and \(9\,\rm{m}\) from the base of the tower are complimentary.

What is Unknown?

To prove height of the tower is \(6\,\rm{m.}\)

Reasoning:

Let the height of the tower as \(CD\). B is a point \(4\rm{m}\) away from the base \(C \) of the tower and \(A\) is a point \(5\rm{m}\) away from the point \(B\) in the same straight line. The angles of elevation of the top \(D\) of the tower from the points \(B\) and \(A\) are complementary.

Since the angles are complementary, if one angle is \(\theta \) and the other is (\(90^\circ \)− \(\theta \)). Trigonometric ratio involving \(CD, \,BC,\, AC\) and angles is \(tan\,\theta \).

Using \(tan\,\theta \) and tan( \(90^\circ \) − \(\theta \)) \(=\) \(cot\,\theta \) ratios are equated to find the height of the tower.

Steps:

In \(\Delta BCD\),

\[\begin{align}& \text{tan}\theta \,\text{= }\frac{CD}{BC} \\ & \text{tan}\theta \,\text{= }\frac{CD}{4}\qquad (1) \\ \end{align}\]

Here,

\[\begin{align}AC&=AB+BC \\ & =5+4 \\ & =9  \end{align}\]

In \(\Delta ACD\),

\[\begin{align}{\rm{tan}}\left( {90 - \theta } \right)\,&=  \frac{{CD}}{{AC}}\\\cot \theta  &= \frac{{CD}}{9}\,\,\,\,\,\,\,\,\qquad\left[ {\,{\rm{tan}}\left( {90 - \theta } \right)\, = \,\cot \theta } \right]\\\frac{1}{{{\rm{tan}}\theta }}\, &= \frac{{CD}}{9}\,\,\,\,\,\,\,\qquad\left[ {\cot \theta  = \frac{1}{{{\rm{tan}}\theta }}} \right]\\{\rm{tan}}\theta  &= \frac{9}{{CD}}\qquad (2)\end{align}\]

From equation (1) and (2)

\[\begin{align}\frac{{CD}}{4} &= \frac{9}{{CD}}\\C{D^2} &= 36\\CD &=  \pm 6\end{align}\]

Since, Height cannot be negative

Therefore, height of the tower is \(6\,\rm{m.}\)

  

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