Question

a round balloon of radius A subtends an angle THETA at the eye of the observer while the angle of elevation of its centre is PHI. prove that the height of the centre of the balloon is AsinPHIcosecTHETA/2

Answer 1

Answer:

Asin∅cosec(θ/2)

Step-by-step explanation:

Hi,

Let the point at which the observer is situated is P

Given that round balloon subtend an angle theta at P,

∠APB = θ

ConsiderΔOAP and ΔOBP, we can observe that

∠OAP = ∠OBP = 90 since PA and PB would be the tangent rays

OP = OP common chord

OA = OB = A radius of balloon

ΔOAP ≅ ΔOBP

Hence ∠APO = ∠BPO

But given that ∠APB = θ = ∠APO + ∠BPO,

Hence ∠APO = ∠BPO = θ/2

Consider ΔOAP,

sin∠APO = OA/OP

⇒sin(θ/2) = A/OP

⇒OP = A/sin(θ/2)

⇒ OP = Acosec(θ/2)

Let F be the foot of the perpendicular from the c3enter of ballon

to the ground, OF = height of the center of the balloon

Given that ∠OPF = ∅

Consider ΔPOF,

sin ∠OPF =  OF/OP

⇒sin∅ = OF/OP

OF = OPsin∅

But OP =  Acosec(θ/2), so

OF =  Asin∅cosec(θ/2)

Height of the center of the balloon is Asin∅cosec(θ/2)

Hope, it helps !

Answer 2

Answer:

BALLOON OF RADIUS = A

POINT OF OBSERVER = O

LET OA AND OB BE THE TANGENTS TO THE BALLOON

HEIGHT = H

ANGLE AT THE EYE OF OBSERVER = θ

ANGLE OF ELEVATION OF ITS CENTRE = Φ

CENTRE = C

IN ΔOAC AND ΔOBC

OA = OB (TANGENTS DRAWN FROM EXTERNAL POINT)

AC = AB (RADII)

OC = OC  (COMMON)

∴ ΔOAB ≅ ΔOAC (BY SSS)

⇒ ∠AOC = ∠BOC

AS, ∠AOB = θ

∴ ∠AOC + ∠BOC = ∠AOB

2∠AOC = ∠AOB

∠AOC = θ/2 = ∠AOB

IN ΔOBC = cosec θ/2 = OC / A

         OC = A cosec θ/2

IN ΔOPC = sin Φ = H / OC

          H = OC sin Φ

       H = A cosec θ/2 sinΦ