Hot air balloon

The center of the balloon is at an altitude of 600 m above the ground (AGL). From habitat on earth is the center of the balloon to see in elevation angle 38°20' and the balloon is seen from the perspective of angle 1°16'. Calculate the diameter of the balloon.

Result

D =  55.507 m

Solution:

h=600 u1=38+20/60=115338.3333 u2=1+16/60=19151.2667 u3=u1+u2=38.3333+1.2667=1985=39.6 x=h/tan(u1 rad)=h/tan(u1 π180 )=600/tan(38.333333333333 3.1415926180 )=758.82371 tan(u2+u1)=(r+h)/x r=x tan(u3 rad)h=x tan(u3 π180 )h=758.82371385966 tan(39.6 3.1415926180 )600=27.75357 D=2 r=2 27.753655.507155.507 mh=600 \ \\ u_{1}=38+20/60=\dfrac{ 115 }{ 3 } \doteq 38.3333 \ \\ u_{2}=1+16/60=\dfrac{ 19 }{ 15 } \doteq 1.2667 \ \\ u_{3}=u_{1}+u_{2}=38.3333+1.2667=\dfrac{ 198 }{ 5 }=39.6 \ \\ x=h / \tan( u_{1} ^\circ \rightarrow\ \text{rad})=h / \tan( u_{1} ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=600 / \tan( 38.333333333333 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=758.82371 \ \\ \tan(u_{2}+u_{1})=(r+h)/x \ \\ r=x \cdot \ \tan( u_{3} ^\circ \rightarrow\ \text{rad})-h=x \cdot \ \tan( u_{3} ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )-h=758.82371385966 \cdot \ \tan( 39.6 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )-600=27.75357 \ \\ D=2 \cdot \ r=2 \cdot \ 27.7536 \doteq 55.5071 \doteq 55.507 \ \text{m}



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