Question

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The
angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After sometime, the angle of elevation reduces to 30º. Find the distance travelled by the balloon during the interval.​

Answer 1

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⎟⎟✪✪ GIVEN QUESTION ✪✪⎟⎟

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. Theangle of elevation of the balloon from the eyes of the girl at any instant is 60°. After sometime, the angle of elevation reduces to 30º. Find the distance travelled by the balloon during the interval.

⎟⎟✪✪ ANSWER ✪✪⎟⎟

✿✿ Refer to Image first ✿✿

◆ Height of balloon from tower=88.2m

◆ Height of the girl = 1.2 m

◆ Angles of elevations = 60° and 30°

◆ Let the distance travelled = d m

From the figure

tan 60° = 87/x

➜ √3 = 87/x

➜ 87 = √3 x ━━━━ᐳ ➀

➜ x = 87/√3 m

Also tan 30° = 87/x+d

➜ 1/√3 = 87/x+d

➜ 87 = x+d/√3 ━━━━ᐳ ➁

From ➀ and ➁

➜ √3x = x+d/√3

➜ √3 ᐧ √3x = x + d

➜ 3x = x + d

➜ 2x =d

Distance travelled = 2 × 87/√3

➜ 2×29×√3×√3 / √3

➜ 58√3 m

The distance travelled by the balloon

during the interval = 58√3 m

Answer 2

HEY HERE IS UR ANSWER :

let the initial position of the balloon be A and final position be B.

Height of balloon above the girl height = 88.2 m - 1.2 m = 87 m

Distance travelled by the balloon =

DE = CE - CD

A/q,

In right ΔBEC,

tan 30° = BE/CE

⇒ 1/√3= 87/CE

⇒ CE = 87√3 m

also,

In right ΔADC,

tan 60° = AD/CD

⇒ √3= 87/CD

⇒ CD = 87/√3 m = 29√3 m

Distance travelled by the balloon = DE = CE - CD = (87√3 - 29√3) m

= 29√3(3 - 1) m

= 58√3 m

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