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  • A spherical gas balloon of radius 16 meter subtends an angle at the eye of the observer A while the angle of elevation of its center from the eye of <img alt="

A spherical gas balloon of radius 16 meter subtends an angle 60^{\circ} at the eye of the observer A while the angle of elevation of its center from the eye of \mathrm{A}$ is $75^{\circ}. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is :
 
Option: 1 8(2+2 \sqrt{3}+\sqrt{2})
Option: 2 8(\sqrt{6}+\sqrt{2}+2)
Option: 3 8(\sqrt{2}+2+\sqrt{3})
Option: 4 8(\sqrt{6}-\sqrt{2}+2)

Answers (1)

best_answer


\frac{PC}{AC}= \sin 30^{\circ}\Rightarrow AC= \frac{PC}{\sin 30^{\circ}}= 2\cdot 16= 32m

and\, \frac{CM}{AC}= \sin \left ( 75^{\circ} \right )= \sin \left ( 45^{\circ}+30^{\circ} \right )
CM= AC\left ( \frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\cdot \frac{1}{2} \right )
CM= 32\left ( \frac{\sqrt{3}+1}{2\sqrt{2}} \right )m= 8\sqrt{2}\left ( \sqrt{3} +1\right )m
\therefore MN= CM+CN= 8\sqrt{6}+8\sqrt{2}+16
                                              \doteq 8\left ( \sqrt{6} +\sqrt{2}+2\right )m

Posted by

Kuldeep Maurya

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