Ex.9.1 Q14 Some Applications of Trigonometry Solution - NCERT Maths Class 10

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Question

A \(1.2 \,\rm{m}\) tall girl spots a balloon moving with the wind in a horizontal line at a height of \(88.2\,\rm{m}\)  from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is \(60^\circ.\) After some time, the angle of elevation reduces to \(30^\circ\) see Figure. Find the distance travelled by the balloon during the interval.

   

Text Solution

  

What is Known?

(i) Height of the girl \(=\) \(1.2\,\rm{m}\)

(ii) Vertical height of balloon from ground \(=\) \(88.2\,\rm{m}\)

(iii) Angle of elevation of the balloon from the eyes of the girl is reducing from \(60^\circ\) to \(30^\circ\) as the balloon moves along wind.

(iv) From the figure, \(OD = BC\) can be calculated as

\(88.2\,\rm{m}\) \(–\) \(1.2\,\rm{m}\) \(=\) \(87\,\rm{m}\dots\)  Result (1)

What is Unknown?

Distance traveled by the balloon, \(OB\)

Reasoning:

Trigonometric ratio involving \(AB,\, BC,\, OD,\,OA\) and angles is tan\(\theta \). [Refer \(AB,\, BC,\, OA\) and \(OD\) from the figure.]

Distance travelled by the balloon \(OB = AB − OA \)

Steps:

From the figure, \(OD = BC\), can be calculated as

    \(88.2\rm {m}\, – 1.2 \rm {m} = 87 {m}\)---------------- (1)

In \(\Delta AOD,\)

\[\begin{align}\tan {{60}^{0}}&=\frac{OD}{OA} \\ \sqrt{3}&=\frac{87}{OA} \\  OA&=\frac{87}{\sqrt{3}} \\ 
&=\frac{87}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\  &=\frac{87\times \sqrt{3}}{3} \\& =29\sqrt{3}\text{m}  \end{align}\]

In \(\Delta ABC,\)

\[\begin{align}\tan {{30}^{0}}&=\frac{BC}{AB} \\ \frac{1}{\sqrt{3}}&=\frac{87}{AB} \\ AB&=87\sqrt{3}  \end{align}\]

Distance traveled by the balloon \(OB = AB \,– OA\)

\[\begin{align}OB  & =87\sqrt{3}-29\sqrt{3} \\ & =58\sqrt{3}  \end{align}\]

Distance traveled by the balloon \(= 58 \sqrt { 3 } \, \,\rm{m}\)