a round balloon of radius alpha subtends an angle teeta at the eye of the observer while the angle of elevation of its centre is beeta. prove that the height of the centre of the balloon is alpha*sin beeta*cosec teeta/2
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heya friend !!!!!111
above is figure
_______________
let us represent the ballon by a circle with centre C and radius r,let PX be the horizontal ground and let O be the point of observation,from P,draw tangent OA and OB yo circle .join CA,CB,And CDperpenducular to OX.
since <AOB=@,. <DOC =b(bita)
<AOC=<BOC=@/2
from right angle ∆OAC,we have .
OC/AA=cosec@/2
=>OC/r=cosec@/2
=>OC=rcosec@/2---------------1)
from right angle ∆ODC,we have
CD/OC=sin b
=>CD=OC*Sin b
=>CD=rsinn bcosec@/2 ------------using 1)
=>hence ,the height of the centre of the ballon from the ground isrsin b cosec@/2
hope it help you
@rajukumar☺☺@@@@
above is figure
_______________
let us represent the ballon by a circle with centre C and radius r,let PX be the horizontal ground and let O be the point of observation,from P,draw tangent OA and OB yo circle .join CA,CB,And CDperpenducular to OX.
since <AOB=@,. <DOC =b(bita)
<AOC=<BOC=@/2
from right angle ∆OAC,we have .
OC/AA=cosec@/2
=>OC/r=cosec@/2
=>OC=rcosec@/2---------------1)
from right angle ∆ODC,we have
CD/OC=sin b
=>CD=OC*Sin b
=>CD=rsinn bcosec@/2 ------------using 1)
=>hence ,the height of the centre of the ballon from the ground isrsin b cosec@/2
hope it help you
@rajukumar☺☺@@@@
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