Keeping the height constant, during the above flight, it was observed that after 15(√3-1) seconds, the angle of elevation changed to 45°. How much is the distance travelled in that duration.

 

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Transcript

Question 38 (ii) – Choice 2 Keeping the height constant, during the above flight, it was observed that after 15(√3 − 1) seconds, the angle of elevation changed to 45°. How much is the distance travelled in that duration.Distance travelled by airplane in 15(√3 − 1) seconds is BY Here, ∠ ABE = 90° & ∠ AYX = 90° In right angle triangle ABE tan A = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴) tan A = (" " 𝐵𝐸)/𝐴𝐵 tan 60° = 𝐵𝐸/𝐴𝐵 √3 = (" " 3000√3)/𝐴𝐵 AB = 3000 m In right angle triangle AXY tan A = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴) tan A = 𝑋𝑌/𝐴𝑌 tan 45° = 𝐶𝐷/𝐴𝑌 1 = (3000√3)/𝐴𝑌 AY = 3000√𝟑 Now, AY = AB + BY 3000√𝟑 = 3000 + BY 3000√𝟑 – 3000 = BY BY = 3000(√𝟑−𝟏)𝒎 Hence, Distance travelled by airplane = 3000(√𝟑−𝟏)𝒎

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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.