Question #adc89

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Answer 1 · 2017-08-14T14:42:56

see below

The problem is depicted in figure below.

enter image source here

I will firstly solve for width of the street #(x)#,

#angle EAD=angle ADB=55^@#.........#(#alternate interior angles#)#

In #Delta ABD#,

#tan55^@=62/x#

#=>x=62/(tan55^@)~~43.413# #m#

Now we see that the height of the building across the street is

#h=CE+ED=CE+62#..........................#(as# #AB=ED)#

Now in #Delta AEC#,

#tan25^@=(CE)/(AE)=(CE)/(x)#

#=>CE=x*tan25^@=62*(tan25^@)/(tan55^@)#

Now,

#h=62*(tan25^@)/(tan55^@)+62~~82.244# #m#