A boy, 1.4 metre tall standing at the edge of a river<br />bank sees the top of a tree on the edge of the other<br />bank at an elevation of 55º. Standing back by 3 metre,<br />he sees it at elevation of 45°<br />(a) Draw a rough figure showing these facts.<br />(b) How wide is the river and how tall is the tree?<br />(sin 55° = 0.8192. cos 55° = 0.5736, tan 55° = 1.4281]
Answers
The height of the tree is 11.39 m and the width of the river is 7 m.
Step-by-step explanation:
Referring to the figure attached below, let’s make some assumptions,
The height of the boy, AB = GC = ED = 1.4 m
BC = 3 m = AG [since the boy stepped 3 m back from his initial position C]
FD = height of the tree
The width of the river = CD = GE
The angle of elevation, ∠FGE = 55°
The angle of elevation after stepping 3 m back from point C, ∠FAE = 45°
Considering ∆FGE, by applying trigonometry ratios of a triangle, we have
tan θ = perpendicular/base = FE/GE
⇒ tan 55° = FE/GE
⇒ 1.4281 = FE/GE
⇒ FE = 1.4281 * GE ……. (i)
Now, considering ∆AFE, by applying trigonometry ratios of a triangle, we have
tan θ = perpendicular/base = FE/AE
⇒ tan 45° = FE/(AG+GE)
⇒ 1 = FE/(3+GE)
⇒ FE = 3 + GE
Substituting value of FE from (i)
⇒ 1.4281GE = 3 + GE
⇒ 0.4281GE = 3
⇒ GE = 3/0.42 81
⇒ GE = 7 m = CD ← Width of the river ……. (ii)
Therefore, from (i) & (ii), we get
The height of the tree,
= FD
= FE + ED
= (1.4281 * GE) + 1.4
= (1.4281 * 7) + 1.4
= 11.39 m
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