Math, asked by Anonymous, 1 year ago

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.


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Answers

Answered by Deepsbhargav
76
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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

 = > theeta \: = \alpha

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)}

______

thus \: the \: hight \: of \: centre \: of \: the \: \\ ballon \: is \: = &gt; \\ <br />= &gt; h \: = \: \frac{r.sin\pi}{sin \frac{\pi}{2} } ............answer

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Answered by ABC007
41
little bit tricky but only follow the steps and do it by ur own
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