The angle of elevation of an airplane from point on the ground is 60°. After a flight of 30 seconds, the angle of elevation becomes 30°. If the airplane is flying at a constant height of 3000√3 m, find the speed of the airplane.
CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard
Question 1 Important
Question 2 Important
Question 3
Question 4 Important
Question 5 Important
Question 6 Important
Question 7
Question 8 Important
Question 9 Important
Question 10
Question 11
Question 12 (OR 1st question)
Question 12 (OR 2nd question)
Question 13 Important
Question 14 Important
Question 15 Important
Question 16
Question 17 Important
Question 18 (OR 1st question) Important
Question 18 (OR 2nd question) Important
Question 19
Question 20
Question 21 Important
Question 22 Important
Question 23 (OR 1st question) Important
Question 23 (OR 2nd question)
Question 24 Important
Question 25 (OR 1st question) Important
Question 25 (OR 2nd question)
Question 26 Important
Question 27 (OR 1st question)
Question 27 (OR 2nd question) Important
Question 28 Important
Question 29 (OR 1st question)
Question 29 (OR 2nd question)
Question 30 Important
Question 31 Important
Question 32 (OR 1st question) Important
Question 32 (OR 2nd question) Important
Question 33 Important
Question 34 Important
Question 35 (OR 1st question) Important
Question 35 (OR 2nd question)
Question 36
Question 37 (OR 1st question)
Question 37 (OR 2nd question)
Question 38 (OR 1st question)
Question 38 (OR 2nd question) Important
Question 39 You are here
Question 40 Important
CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard
Last updated at Nov. 1, 2019 by Teachoo
Question 39 The angle of elevation of an airplane from point on the ground is 60°. After a flight of 30 seconds, the angle of elevation becomes 30°. If the airplane is flying at a constant height of 3000√3 m, find the speed of the airplane. Let the point on ground be A Since airplane is flying at height of 3000√3 m CD = BE = 3000√3 m We need to find speed of airplane To find speed, we need to distance travelled And distance travelled by airplane in 30 seconds is BC So, we need to find BC Let BC = x Here, ∠ ABE = 90° & ∠ ACD = 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 ACD tan A = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴) tan A = 𝐶𝐷/𝐴𝐶 tan 30° = 𝐶𝐷/𝐴𝐶 1/√3 = (3000√3)/𝐴𝐶 AC = 3000√3 × √3 = 9000 m Now, AC = AB + BC 9000 = 3000 + BC 9000 – 3000 = BC BC = 9000 – 3000 BC = 6000 m Hence, Distance travelled by airplane in 30 seconds = 6000 m Now, Speed of airplane = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑇𝑖𝑚𝑒 = (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑖𝑛 30 𝑠𝑒𝑐𝑜𝑛𝑑𝑠)/(30 𝑠𝑒𝑐𝑜𝑛𝑑𝑠) = 6000/30 m/s = 200 m/s