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Chapter 9 Class 10 Some Applications of Trigonometry
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### Transcript

Ex 9.1 , 14 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 60°. After some time, the angle of elevation reduces to 30° (see figure ). Find the distance travelled by the balloon during the interval. Given that 1.2 m tall girl sees a balloon So, AG = 1.2 m Also, AG & BF are parallel BF = AG = 1.2 m And the balloon is at a height of 88.2 m So, EF = 88.2 Girl sees balloon first at 60° So, ∠ EAB = 60° After travelling, the angle of elevation becomes 30° So, ∠ DAC = 30° Distance travelled by the balloon = BC We have to find BC Now, BE = EF – BF BE = 88.2 – 1.2 m BE = 87 m Also, BE = DC = 87 m Here, ∠ ABE = 90° & ∠ ACD = 90° Now, AC = AB + BC 87√3 = (" " 87)/√3 + BC 87√3 – (" " 87)/√3 = BC BC = 87√3 – (" " 87)/√3 BC = 87 (√3 – (" " 1)/√3 ) BC = 87((√3 √3 −1)/√3) BC = 87((3 − 1)/√3) BC = 87 (2/√3) BC = (" " 87 ×2)/√3 Multiply √3 in numerator and denominator BC = (" " 87 × 2)/√3 × √3/√3 BC = (" " 87 × 2 ×√3)/3 BC = 29×2×√3 BC = 58√3 Hence, distance travelled by balloon = 58√3 m

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.