Question

a boy standing at the edge of canal see the top of tree at the other bank at an angle of elevation 54°. stepping back 20m, he sees the top of treeat an elevation 27°
draw rough figure
?find the height of the tree
find the width of canal​

Answer 1

Step-by-step explanation:

Let CD=h be the height of the tree and BC=x be the breadth of the river.

From the figure ∠DAC=30

∘

and ∠DBC=60

∘

In right angled triangle △BCD,tan60

∘

=

BC

DC

⇒

3

=

x

h

⇒h=x

3

.....(1)

From the right-angled triangle △ACD

tan30

∘

=

40+x

h

⇒

3

1

=

40+x

h

⇒

3

h=40+x .......(2)

From (1) and (2) we have

3

(x

3

)=40+x

⇒3x=40+x

⇒3x−x=40

⇒2x=40

⇒x=

2

40

=20

From (1) we get h=x

3

=20

3

=20×1.732=34.64m

∴ Height of the tree=34.64 m and width of the river=20m

Answer 2

Given:

Angle of elevation when boy is at edge of canal bank = 54°

Angle of elevation when he moves 20m back = 27°

To find:

Height of tree.

Width of canal.

Solution:

A rough figure is shown below.

Let height of tree be $h$ and let the width of canal be $(x-20)m$ where $x$ is the total distance BC. Let B be a point on the other bank.

The tree can be considered as ideally perpendicular to the ground thus making 90° with the ground.

In right angle ΔABD,

$Tan 54=\frac{AB}{BD}$

$1.37=\frac{h}{x-20}$

$h=1.37(x-20)$

$h=1.37x-27.4...(1)$

In right angled ΔABC,

$Tan27=\frac{AB}{BC}$

$\frac{1}{2}=\frac{h}{x}$

$x=2h...(2)$

Substituting equation (2) in equation (1)

$x=2(1.37x-27.4)$

$x=2.74x-54.8$

$54.8=2.74x-x$

$54.8=1.74x$

$x=\frac{54.8}{1.74}=31.5m$

Hence, the width of canal is $x-20=31.5-20=11.5m$

Substituting the value of $x$ in equation (2)

$x=2h$

$h=\frac{x}{2}$

$h=\frac{31.5}{2}=15.75m$

Hence, the height of tree is $15.75m$.

The height of tree is $15.75m$ and the width of canal is $11.5m$.