Question

A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on opposite bank is 60°. When he moves 30 m away from the bank, he finds the angle of elevation to be 30°. Find the height of the tree and width of the river. [Take `sqrt(3)` = 1.732]

Answer 1

Let LCD be the tree of height h m. Let B be the position bank of the river After moving 30 m away from point B.

Let new position of man be A i.e., AB = 30 m. The angles of elevation of the top of the tree from points A and B are 30° and 60° respectively, i.e., ∠CAD = 30° and ∠CBD = 60°

Let BC = x m

In right triangle BCD, we have

`tan 60^circ = "CD"/"BC"`

⇒ `sqrt(3) = h/x`

⇒ `x = h/sqrt(3)`   ...(i)

In right triangle ACD, we have

`tan 30^circ = "CD"/"AC"`

⇒ `1/sqrt(3) = h/(x + 30)`

⇒ `x + 30 = sqrt(3)h`

⇒ `x = sqrt(3)h - 30`   ...(ii)

Comparing equations (i) and (ii), we get

⇒ `h/sqrt(3) = sqrt(3)h - 30`

⇒ `h = 3h - 30sqrt(3)`

⇒ `-2h = -30sqrt(3)`

⇒ `h = 15sqrt(3)`

= 15 × 1.732

= 25.98 m

Hence, the height of the tree is 25.98 m.

Now, substituting the value of `h = 15sqrt(3)` in equation we get,

`x = h/sqrt(3)`

⇒ `x = (15sqrt(3))/sqrt(3)`

⇒ x = 15 m

Hence, width of the river is 15 m.