Question

A person standing on the bank of river observes that the angle of the elevation of the top of a tree standing on the opposite bank is 60 degree when he moves 40m away from bank he finds the angle of elevation to be 30 degree find the height of the tree and width of the canal

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given:

Angle of elevation = 60

Distance = 40m

New angle of elevation = 30

To Find:

Height of the tree and width of the canal

Solution:

Let the breadth of the river = BC = x

Let height of the tree = CD = h

Now, ∠DAC=30  and ∠DBC=60

Thus, In Δ BCD

Tan 60 = DC/BC

= √3 = h/x

h = x√3 --- eq 1

In Δ ACD

Tan30 = h / 40 + x

1/√3  = h / 40 + x

= √3h = 40 + x --- eq 2

From eq 1 and 2

√3 ( x√3) = 40 + x

3x = 40 + x

2x = 40

x = 40/2

x = 20

Thus,

x√3 = 20√3

= 20 x 1.73

= 34.6

Answer: Height of the tree is 34.64 m and width of the canal is 20m