Question

A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank is 60°. When he moves 40 metres away from the bank, he finds the angle of elevation to be 30°. Find the height of the tree and the width of the river.​

Answer 1

Answer:

$\huge{\underline{\mathfrak{\pink{Answer:-}}}}$

__________________

Step by step explanation:-

______________

According to figure ;

___________

let DC be height and BC be the width

Given:-

AB =40m

<DBC = 60°

<DAC =30°

_____________

To find :-

DC =?(height)

BC =?(width)

_____________

Solution:-

In right∆ ACD:-

(using relation of Tanθ)

Tan θ= P/B

Tan 30°= DC/AC

1/√3= h/AC

1/√3 =h/AB+BC

(AC=AB+BC)

1/√3 = h/40+x

√3h=40+x. (eq1)

Now,In right ∆DCB

(using relation of Tanθ)

Tan θ = P/B

Tan 60°=DC/BC

√3= h/x

h = √3x. (equation 2)

___________

Now put the value of h in equation 1

√3h = 40+x

√3(√3x)=40+x

3x= 40+x

3x-x =40

2x=40

x=40/2 =20m

_____________

Now put the value of x in equation2

h =√3x

h=√3(20)

h=20√3m

_______________________

Since ,

the height of tree =20√3m

Width of river = 20m

________________

Hence we get the answer ;

____________

$\huge\mathfrak\pink{Thanks}$

Be brainly⭐

Answer 2

Solution:

=> Let height of the tree be h m.

=> And, width of river be x m.

=> C and D are final and initial point of the man.

Now, In ΔABD,

$\sf{\implies \tan\;30^{\circ}=\dfrac{AB}{BD}}$

$\sf{\implies \dfrac{1}{\sqrt{3}}=\dfrac{h}{x+40}}$

$\sf{\implies h = \dfrac{x+40}{\sqrt{3}}\;\;\;\;........(1)}$

Now, In ΔABC,

$\sf{\implies \tan 60^{\circ}=\dfrac{AB}{BC}}$

$\sf{\implies \sqrt{3} =\dfrac{h}{x}}$

Now, put the value of h from Eq (1)

$\sf{\implies \sqrt{3} =\dfrac{x+40}{\sqrt{3}x}}$

$\sf{\implies 3x = x+40}$

$\sf{\implies 2x = 40}$

$\sf{\implies x = \dfrac{40}{2}}$

$\sf{\implies x = 20\;m}$

∴ Width of river = 20 m

Now, put the value of x in Eq (1) to find height if tree.

$\sf{\implies h = \dfrac{x+40}{\sqrt{3}}}$

$\sf{\implies h = \dfrac{20+40}{\sqrt{3}}}$

$\sf{\implies h=\dfrac{60}{\sqrt{3}}}$

$\sf{\implies h = \dfrac{60}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}}$

$\sf{\implies h = \dfrac{60\sqrt{3}}{3}}$

$\sf{\implies h=20\sqrt{3}}$

∴ Height of tree is 20√3 m.