Question

The two palm trees are of equal heights and are standing opposite each
other on either side of the river, which is 80 m wide. From a point O
between them on the river the angles of elevation of the top of the trees
are 60° and 30°, respectively. Find the height of the trees and the
distances of the point O from the trees.
Need urgently no rubish answer required

Answer 1

Answer:

Let BD=river

AB=CD=palm trees=h

BO=x

OD=80-x

In ∆ABO,

Tan60˚=h/x

√3=h/x -----------------------(1)

H=√3x

In ∆CDO,

Tan 30˚=h/(80-x)

1/√3= h/(80-x) ---------------------(2)

Solving (1) and (2), we get

X=20

H=√3x=34.6

the height of the trees=h=34.6m

BO=x=20m

DO=80-x=80-20=60m

Answer 2

Given:

Distance between two trees=80m

To find:

i. The height of the trees

ii. The distance of point O from the trees

Solution:

The height of the trees is 20$\sqrt{3}$ m and the distance of point O from the trees is 60 m and 20 m.

We can find the solution by following the given steps-

We know that the height and distance can be calculated by using trigonometry.

Let the height of the trees be H and the distance from O to one of the trees is D.

The distance between O and the other tree=(80-D)

Now, point O is making an angle of 30° and 60° with each tree.

Tan theta=Perpendicular/Base

So, tan 30°=H/D

$1/\sqrt{3}$=H/D

D/$\sqrt{3}$=H (1)

Similarly, tan 60°=H/(80-D)

$\sqrt{3}$=H/(80-D)

$\sqrt{3}$ (80-D)=H

From (1),

$\sqrt{3}$(80-D)=D/$\sqrt{3}$

3(80-D)=D

240-3D=D

240=4D

D=60 m

The distance between O and other tree=(80-D)=80-60=20m

So, H=60/$\sqrt{3}$

H=20$\sqrt{3}$ m

Therefore, the height of the trees is 20$\sqrt{3}$ m and the distance of point O from the trees is 60 m and 20 m.