A spherical balloon if radius r subtends an angle theta at the eye of an observer. If the angle of elevation of its centre is pie, find the height of the centre of the balloon.
Rahul053:
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HERE IS YOUR ANSWER ☞
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let's the hight of centre of the balloon above the ground be = h meter..
since the balloon subtends an angle at = theeta
now suppose that
Now
=> angle(EAD) = alpha
in △ACE = △ACD
=> AE = AD... [ lengths of tangents drawn from an external point to the circle are equal ]
=> AC = AC.... [ common ]
=> angle(CEA) = angle(CDA) = 90 ...[ radius is perpendicular to tangent at point of contact ]
=> △ACE. ≈ △ACD
=> angle(EAC) = angle(DAC)...[C.P.C.T ]
=> angle(EAC) = angle(DAC) = alpha/2
________
in right △ACD
=> sin(altha/2) = CD/AC
=> AC = r/sin(alpha/2) _____eq(1)
________
Now in right △ABC
=> sin(pai) = CB/AC
=> CB = AC×sin(pai)
=> CB = [ r/sin(alpha/2)].sin(pai)___{by eq(1)}
______
=========================
HOPE IT WILL HELP YOU ☺☺
=========================
DEVIL_KING ▄︻̷̿┻̿═━一
HERE IS YOUR ANSWER ☞
=====================
let's the hight of centre of the balloon above the ground be = h meter..
since the balloon subtends an angle at = theeta
now suppose that
Now
=> angle(EAD) = alpha
in △ACE = △ACD
=> AE = AD... [ lengths of tangents drawn from an external point to the circle are equal ]
=> AC = AC.... [ common ]
=> angle(CEA) = angle(CDA) = 90 ...[ radius is perpendicular to tangent at point of contact ]
=> △ACE. ≈ △ACD
=> angle(EAC) = angle(DAC)...[C.P.C.T ]
=> angle(EAC) = angle(DAC) = alpha/2
________
in right △ACD
=> sin(altha/2) = CD/AC
=> AC = r/sin(alpha/2) _____eq(1)
________
Now in right △ABC
=> sin(pai) = CB/AC
=> CB = AC×sin(pai)
=> CB = [ r/sin(alpha/2)].sin(pai)___{by eq(1)}
______
=========================
HOPE IT WILL HELP YOU ☺☺
=========================
DEVIL_KING ▄︻̷̿┻̿═━一
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41
little bit tricky but only follow the steps and do it by ur own
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