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Answers
Step-by-step explanation:
Given, balloon subtends α angle at the observers eye.
Let the height of the center of the balloon be h m.
It is given that,
∠AOB = α, ∠DOC = β.
⇒ OA = OB {Length of tangents drawn from external point are equal}
⇒ OC = OC {common}
⇒ CA = CB {radius of circle}
∴ OCA = OCB(SSS congruence criterion}
⇒ ∠AOC = ∠BOC = α/2
(i) In ΔOAC:
cosec α/2 = OC/AC
⇒ cosec α/2 = OC/r
⇒ OC = r. cosec α/2
(ii) In ΔODC:
sinβ = h/OC
⇒ h = OC.sinβ
Substitute (i) in (ii), we get
⇒ h = r. cosecα/2. sinβ
⇒ h = r sinβ cosecα/2
Hope it helps!
round baloon with centre O & radius r subtends, angle £ at point E. £ is bisected into £/2 by EO. EA & EB are tangents , which will be perpendicular to OA & OB respectively.
Angle of elevation of O at E = β
TO FIND : Height of the centre of the baloon from the ground = h
In triangle OAE,
Sin £/2 = r / OE
=> OE = r/ Sin £/2 …………..(1)
In triangle OCE,
Sin β = OC / OE
=> Sin β = h/(r/ Sin £/2)
=> Sin β = (h Sin £/2 ) /r
=> r Sin β = h Sin £/2
=> h = r * Sin β * cosec £/2