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)

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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

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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