Answer:
Let DC be the tower and BC be the building. Then,
∠CAB=45
o
,∠DAB=60
,BC=20 m
Let height of the tower, DC=h m.
In right △ABC,
tan45 o = AB/BC
1=20/AB
AB=20m
In right∆ABD,
tan 60°=BD/AB
√3=h+20/20
h=20(√3-1)m
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