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There is a math problem that I can't do. I know the answer, but I don't know how to use it I want to ask you.
Use the knowledge that the height of an object at the same time of the day is directly proportional to the shadow growth.

Solution: Make the vertical line BC of DE in E and extend the intersection BC of AD in F. 。

∠ DCE = 30, then DE = CD/2 = 2;; CE=√(CD? De? )=2√3.

∫DE/EF = GH/HP ..

That is 2/EF= 1/2.

∴ ef = 4m。 Then BF=BC+CE+EF= 14+2√3.

AB/BF=GH/HP= 1/2, then AB=BF/2=(7+√3) meters.