Question

The angle of elevation of the top of a tower, 300 m high, from a point on the ground is observed as 30°. At an instant a hot air balloon passes vertically above the tower and at that instant its angle of elevation from the same point on the ground is 60°. Find the height of the balloon from the ground and the distance of the tower from the point of observation. (Use `sqrt(3)` = 1.73)

Answer 1

Let BD be the height of the tower and the height of the balloon from the top of the tower is CD.

Height of the tower (BD) = 300 m

Let height of the balloon from top of the tower (CD) = h

Height of the balloon from ground (BC) = h + 300 m

Distance between from point A and tower (AB) = x

Angle of elevation from point A to the top of the tower = 30°

So, ∠DAB = 30°

Angle of elevation from point A to the top of the balloon = 60°

So, ∠CAB = 60°

In right angle triangle DAB,

⇒ tan 30° = `(BD)/(AB)`

⇒ `1/sqrt(3) = 300/x`   ...`(∵ tan 30^circ = 1/sqrt(3))`

⇒ x = `300sqrt(3)`   ...(1)

⇒ x = 300 × 1.73

⇒ x = 519 m

In right angle triangle ABC,

⇒ tan 60° = `(BC)/(AB)`

⇒ `sqrt(3) = (BD + CD)/(AB)`

⇒ `sqrt(3) = (300 + h)/x`   ...(2)

Now, putting value of x in equation (2),

`sqrt(3) = (300 + h)/(300sqrt(3))`

⇒ `sqrt(3) xx 300sqrt(3) = 300 + h`   ...(Cross multiplying)

⇒ 900 = 300 + h

⇒ h = 900 – 300

⇒ h = 600 m

Height of balloon from ground (BC) = BD + CD

= 300 + 600

= 900 m

Hence, height of the balloon from ground is 900 m and distance between point A and tower AB is 519 m.