a child is watching a bird on a tree. the angle of elevation of his/her sight is 45° . the child is 1.5 m tall. find the height of the tree when the distance between the foot of the tree and child on the ground is 30 m.
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BF represents the tree, DG is the first position of the boy and AE is the new position.
CD represents the river.
We have : AE = DG
Let DC = x, then EC = (x + 3) m.
tan45° = EC/CF
⇒1=CF /x+3
⇒CF=x+3 ...... (i)
tan55° = CF/DC
⇒1.4281= CF/x
⇒CF=1.4281×x ..... (ii)
From (i) and (ii), we have
1.4281x=x+3
0.4281x=3⇒x= 0.42813 ∼7m
Width of the river = CD = 7 m
Height of the tree = BF = BC = CF = (1.4 + 7 + 3)m = 11.4
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