Q) One observer estimates the angle of elevation to the basket of a hot air balloon to be 60°, while another observer 100 m away estimates the angle of elevation to be 30°. Find:
(a) The height of the basket from the ground.
(b) The distance of the basket from the first observer’s eye.
(c) The horizontal distance of the second observer from the basket.
Important Note: Generally, it is given in the question that the two observers are on same side of the balloon or on opposite sides. In such a case (like this question) where it is not given, same side will be considered. This is important to note, Students. In case you take opposite sides for this question; your all three answers will be completely different and you will loose all marks.
Let’s draw the diagram keeping both observers on same side and solve the question:
Let the hot air balloon be at point B, C is the1st observer with ∠ BCA = 600, D is the 2nd observer with ∠ BDA = 300, given that distance CD = 100 m.
We need to find out, H – height of balloon; S – distance of 1st observer C from Balloon B and D – horizontal distance of 2nd observer D from basket.
Let’s start from Δ ABD,
tan 30° =
or D = H√3 ………… (i)
Next in Δ ABC, tan 60° =
or √3 =
or H = √3 (D – 100)
Substituting value of D from equation (i), we get:
H = √3 (H√3 – 100)
H = 3H – 100√3
H = 50 √3
Therefore, height of the balloon is 50√3 m. …….. Ans (a)
Next in Δ ABC, sin 60° =
or S =
Since H = 50√3
= 100 m
Therefore, the distance of 1st observer from balloon is 100 m…… Ans (b)
From equation (i), we get, D = H√3 = (50√3) (√3) = 150
Therefore, horizontal distance of 2nd observer from balloon is 150 m…… Ans (c)