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Q14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60º. After some time, the angle of elevation reduces to 30º. Find the distance travelled by the balloon during the interval.


Answer :


Let P and Q be the two positions of the balloon and let A be the point of observation. let ABC be the horizontal through A. It is given that angles of elevation of the balloon in two position P and Q from \( ∠ \ PAB \ = \ 60º \) , \( ∠ \ QAB \ = \ 30º \). It is also given that MQ = 88.2

\( => \ CQ \ = \ MQ \ - \ MC \ = \ 88.2 \ - \ 1.2 \ = \ 87 \)m .



In \( ∆ \ ABP \), we have

\( \frac{BP}{AB} \ = \ tan60º \ => \ \frac{87}{AB} \ = \ \sqrt{3} \) \( => \ AB \ = \ \frac{87}{ \sqrt{3}} \ = \ \frac{87 \sqrt{3}}{3} \ = \ 29 \sqrt{3} \)      (1)

In \( ∆ \ ACQ \), we have

\( \frac{CQ}{AC} \ = \ tan30º \ => \ \frac{87}{AC} \ = \ \frac{1}{ \sqrt{3}} \) \( => \ AC \ = \ 87 \sqrt{3} \)      (2)

∵ \( PQ \ = \ BC \ = \ AC \ - \ AB \ = \ 87 \sqrt{3} \ - \ 29 \sqrt{3} \ = \ 58 \sqrt{3} \) \( = \ 100.456 \)m

∴ the balloon travels \( 100.456 \)m