Ex.9.1 Q14 Some Applications of Trigonometry Solution - NCERT Maths Class 10
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 \(= 58 \sqrt { 3 } \, \,\rm{m}\)