# Clouds

From two points A and B on the horizontal plane was observed forehead cloud above the two points under elevation angle 73°20' and 64°40'. Points A , B are separated by 2830 m. How high is the cloud?

Result

x =  3662.055 km

#### Solution:

$u_{ 1 } = 73+20/60 = \dfrac{ 220 }{ 3 } \doteq 73.3333 \ \\ u_{ 2 } = 64+40/60 = \dfrac{ 194 }{ 3 } \doteq 64.6667 \ \\ t_{ 1 } = \tan( u_{ 1 } ^\circ \rightarrow rad) = \tan( u_{ 1 } ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ ) = \tan( 73.333333333333 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ ) = 3.34023 \ \\ t_{ 2 } = \tan( u_{ 2 } ^\circ \rightarrow rad) = \tan( u_{ 2 } ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ ) = \tan( 64.666666666667 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ ) = 2.11233 \ \\ s = 2830 \ \\ a+b = s \ \\ x = a \cdot \ t_{ 1 } = b \cdot \ t_{ 2 } \ \\ a \cdot \ t_{ 1 } = (s-a) \cdot \ t_{ 2 } \ \\ a = s \cdot \ t_{ 2 }/(t_{ 1 }+t_{ 2 }) = 2830 \cdot \ 2.1123/(3.3402+2.1123) \doteq 1096.3473 \ \\ b = s-a = 2830-1096.3473 \doteq 1733.6527 \ \\ x = a \cdot \ t_{ 1 } = 1096.3473 \cdot \ 3.3402 \doteq 3662.055 = 3662.055 \ \text{ km }$

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