# Clouds

Approximately at what height is the cloud we see under an angle of 26°10' and see the Sun at an angle of 29°15' and the shade of the cloud is 92 meters away from us?

Result

h =  366.759 m

#### Solution:

$A=26 + 10/60=\dfrac{ 157 }{ 6 } \doteq 26.1667 \ ^\circ \ \\ B=29 + 16/60=\dfrac{ 439 }{ 15 } \doteq 29.2667 \ ^\circ \ \\ l=92 \ \text{m} \ \\ h=x \cdot \ \tan(B)=366.759 \ \text{m} \ \\ h=(x+l) \cdot \ \tan(A) \ \\ x \cdot \ \tan(B)=(x+l) \cdot \ \tan(A) \ \\ x \cdot \ \tan(B)=x \cdot \ \tan(A) + l \cdot \ \tan(A) \ \\ x=l \cdot \ \tan A ^\circ / (\tan B ^\circ - \tan A ^\circ )=l \cdot \ \tan 26.1666666667^\circ \ / (\tan 26.1666666667^\circ \ - \tan 26.1666666667^\circ \ )=92 \cdot \ \tan 26.1666666667^\circ \ / (\tan 26.1666666667^\circ \ - \tan 26.1666666667^\circ \ )=l \cdot \ 0.491339 / (0.491339 - 0.491339)=654.44923 \ \\ h=x \cdot \ \tan B ^\circ =x \cdot \ \tan 29.2666666667^\circ \ =654.449231886 \cdot \ \tan 29.2666666667^\circ \ =x \cdot \ 0.560409=366.7593 \ \\ h_{2}=(x+l) \cdot \ \tan A ^\circ =(x+l) \cdot \ \tan 26.1666666667^\circ \ =(654.449231886+92) \cdot \ \tan 26.1666666667^\circ \ =(x+l) \cdot \ 0.491339=366.7593$

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