Question

tower height 25m angle of elevation 30 angle of depression 60 what is the height of the building

Answer 1

In ∆ABC
tan30°=AC/BC

$\frac{1}{ \sqrt{3} }$
=h-H/x
x=√3(h-H).....i

in∆BED
tan60°=25/x
$\tan(60) = \sqrt{3}$
√3=25/x
x=25/√3
x=25√3/3(by rationalising with√3)....ii

equating I and ii
√3(h-H)=25√3/3
H=25
h=100/3

height of building=h
h=100/3