**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.

**Ans: **

**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 = 60^{0}, D is the 2nd observer with ∠ BDA = 30^{0}, 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

or S =

Since H = 50√3

= 100 m

**Therefore, the distance of 1 ^{st }observer from balloon is 100 m**

**…… Ans (b)**

From equation (i), we get, D = H√3 = (50√3) (√3) = 150

**Therefore, horizontal distance of 2 ^{nd} observer from balloon is 150 m…… Ans (c)**