Question 34 - CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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Question 34 A boy whose eye level is 1.35 m from the ground, spots a balloon moving with the wind in a horizontal line at some height from the ground. The angle of elevation of the balloon from the eyes of the boy at an instant is 60°. After 12 seconds, the angle of elevation reduces to 30°. If the speed of the wind is 3m/s then find the height of the balloon from the ground. (Use √3=1.73)Given that 1.35 m tall boy sees a balloon So, AF = 1.35 m Also, AF, BG & CH are parallel BG = CH = AF = 1.35 m Also, given that After 12 seconds, the angle of elevation reduces to 30°. If the speed of the wind is 3m/s then find the height of the balloon from the ground. Therefore, BC = Distance travelled by balloon due to 3m/s wind in 12 sec = Speed × Time = 3 × 12 = 36 m ∴ BC = 36 m And, we need to find the height of the balloon from the ground. i..e we need to find DH Also, boy sees balloon first at 60° So, ∠ EAB = 60° After travelling, the angle of elevation becomes 30° So, ∠ DAC = 30° And EG = DH because height of balloon from ground remains same Here, ∠ ABE = 90° & ∠ ACD = 90° In right angle triangle EBA tan A = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴) tan A = (" " 𝐵𝐸)/𝐴𝐵 tan 60° = 𝐵𝐸/𝐴𝐵 √3 = (" " 𝐵𝐸)/𝐴𝐵 AB = 𝑩𝑬/√𝟑 In right angle triangle DAC tan A= (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴) tan A = (" " 𝐶𝐷)/𝐴𝐶 tan 30° = 𝐶𝐷/𝐴𝐶 1/√3 = 𝐶𝐷/𝐴𝐶 AC = CD√𝟑 Since BE = CD ∴ AB = 𝑪𝑫/√𝟑 & AC = CD√𝟑 Now, BC = AC − AB 36 = 𝐂𝐃√𝟑−𝑪𝑫/√𝟑 36 = CD(√3−1/√3) 36 = CD((√3 ×√3 − 1 )/√3) 36 = CD((3 − 1 )/√3) 36 = CD(2/√3) 36 × √3/2=𝐶𝐷 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 18√3=𝐶𝐷 CD=18√3 CD = 18 × 1.73 CD = 31.14 m Now, Height of balloon = DH = CD + CH = 31.14 + 1.35 = 32.49 m