Question

A boy, 1.4 metres tall standing at the edge of a river bank sees the top of a tree on the edge of the other bank at an elevation of 55°. Standing back by 3 metres, he sees it at elevation of 45°. (a) Draw a rough figure showing these facts. (b) How wide is the river and how tall is the tree? [ sin 55° = 0.8192, cos 55° = 0.5736, tan 55° = 1.4281]

Answer 1

The width of the river = 7 m

The height of the tree = 11.4 m

Step-by-step explanation:

AC is the height of the tree

DE is the first position of the boy

GF is the another position of the boy

GF=DE

let

DB = x

GB = X+3

then,

tan 45° =$\frac{AB}{GB}$

1= $\frac{AB}{x+3}$

AB = x+3

similarly

tan 55° = $\frac{AB}{DB}$

using tan 55° = 1.4281

1.4281 = $\frac{AB}{x}$

AB= 1.4281(x)

using equation (1) and (2)

1.4281(x) = x+3

x= 7 m

putting the value of x in eq (1)

AB = x+3

AB = 7+3 = 10 m

then

The height of the tree = AB+BC

= 10 + 1.4 = 11.4 m

hence ,

The width of the river = 7 m

The height of the tree = 11.4 m

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Answer 2

Answer:

width of the river equal to CD equal to 7 centimetre height of equal to b + BC + CA is equal to 11.4 metre