Mast has 13 m long shadow on a slope rising from the mast foot in the direction of the shadow angle at angle 13.3°. Determine the height of the mast, if the sun above the horizon is at angle 45°12'.


v =  17.65 m


l=13 m A=13.3  B=45.2   sin(A)=hl h=l sinA=l sin13.3 =13 sin13.3 =l 0.23005=2.99065  x=l cosA=l cos13.3 =13 cos13.3 =l 0.973179=12.65133  tan(A+B)=h+vx v=x tan((A+B) rad)h=x tan((A+B) π180 )h=12.6513253461 tan((13.3+45.2) 3.1415926180 )2.99064658345=17.65444=17.65 ml=13 \ \text{m} \ \\ A=13.3 \ ^\circ \ \\ B=45.2 \ ^\circ \ \\ \ \\ \sin(A)=\dfrac{ h }{ l } \ \\ h=l \cdot \ \sin A ^\circ =l \cdot \ \sin 13.3^\circ \ =13 \cdot \ \sin 13.3^\circ \ =l \cdot \ 0.23005=2.99065 \ \\ \ \\ x=l \cdot \ \cos A ^\circ =l \cdot \ \cos 13.3^\circ \ =13 \cdot \ \cos 13.3^\circ \ =l \cdot \ 0.973179=12.65133 \ \\ \ \\ \tan(A+B)=\dfrac{ h+v }{ x } \ \\ v=x \cdot \ \tan(( A+B) ^\circ \rightarrow\ \text{rad}) - h=x \cdot \ \tan(( A+B )^\circ \cdot \ \dfrac{ \pi }{ 180 } \ ) - h=12.6513253461 \cdot \ \tan(( 13.3+45.2 )^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ ) - 2.99064658345=17.65444=17.65 \ \text{m}

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