Question 1204257
<font color=black size=3>
57° 20' = 57 + (20/60) = 57.33333° approximately
18° 10' = 18 + (10/60) = 18.16667° approximately


Diagram
*[illustration Screenshot_317.png]


The person's eye is located at point A.
B is just across from A.
C and D are the top and base of the other building.


Focus on triangle ABC.
tan(angle) = opposite/adjacent
tan(angle CAB) = BC/AB
tan(57.33333°) = x/38
x = 38*tan(57.33333°)


Now turn your focus to triangle ABD.
tan(angle) = opposite/adjacent
tan(angle DAB) = BD/AB
tan(18.16667°) = y/38
y = 38*tan(18.16667°)


The total height of CD is
CD = CB + BD
CD = x + y
CD = 38*tan(57.33333°) + 38*tan(18.16667°)
CD = 38*( tan(57.33333°) + tan(18.16667°) )
CD = 71.73617


The building is roughly 71.736 meters tall.
Round this approximate value however your teacher instructs.
</font>