Question

a person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60° when he moves 200m away from the bank he finds that the angle of elevation to be 30° calculate width of the river and height of the tree​

Answer 1

Answer :

• The height of the tree , h = 100√3.

• The width of the river , x = 100 m.

Explanation :

Given :

• Angle of elevation = ∠1 = 60°.

• Angle of elevation = ∠2 = 30°.

• Distance moved by the person = 200 m.

To find :

• Height of the tree , h = ?

• Width of the river , x = ?

Solution :

According to the diagram ∆CBO ,

• Height of the triangle is CB.

• Base of the triangle is OB.

Let the base be x m and height be h m.

Since we have to find the height and case , we will use tan θ , since ;

$\boxed{\bf{tan\theta = \dfrac{h}{b}}}$

Where :

• h = Height
• b = Base

Now using the tan θ and substituting the values in it, we get :

$:\implies \bf{tan\theta = \dfrac{h}{b}}$

$:\implies \bf{tan60^{\circ} = \dfrac{h}{b}}$

$:\implies \bf{\sqrt{3} = \dfrac{h}{x}}\quad [\because \bf{tan60^{\circ} = \sqrt{3}}]$

$:\implies \bf{x = \dfrac{h}{\sqrt{3}}}$

$\boxed{\therefore \bf{x = \dfrac{h}{\sqrt{3}}}}$

Hence the width of the river in terms of h is h/√3.

Now, let's consider ∆ABC , from there we get :

• CB = Height
• AB = Base

By using tan θ and substituting the values in it, we get :

$:\implies \bf{tan\theta = \dfrac{h}{b}}$

$:\implies \bf{tan30^{\circ} = \dfrac{h}{200 + x}}$

$:\implies \bf{tan30^{\circ} = \dfrac{h}{200 + \dfrac{h}{\sqrt{3}}}}\quad [\because \bf{x = \dfrac{h}{\sqrt{3}}}]$

$:\implies \bf{tan30^{\circ} = \dfrac{h}{\dfrac{200\sqrt{3} + h}{\sqrt{3}}}}$

$:\implies \bf{tan30^{\circ} = \dfrac{h}{200\sqrt{3} + h} \times \sqrt{3}}$

$:\implies \bf{tan30^{\circ} = \dfrac{h\sqrt{3}}{200\sqrt{3} + h}}$

$:\implies \bf{\dfrac{1}{\sqrt{3}} = \dfrac{h\sqrt{3}}{200\sqrt{3} + h}}\quad [\because \bf{tan30^{\circ} = \dfrac{1}{\sqrt{3}}}]$

$:\implies \bf{200\sqrt{3} + h = h\sqrt{3} \times \sqrt{3}}$

$:\implies \bf{200\sqrt{3} + h = 3h}$

$:\implies \bf{200\sqrt{3} = 3h - h}$

$:\implies \bf{200\sqrt{3} = 2h}$

$:\implies \bf{\dfrac{200\sqrt{3}}{2} = h}$

$:\implies \bf{100\sqrt{3} = h}$

$\boxed{\therefore \bf{h = 100\sqrt{3}\:.}}$

Now , substituting the value of h in the equation of x , we get :

$:\implies \bf{x = \dfrac{h}{\sqrt{3}}}$

$:\implies \bf{x = \dfrac{100\sqrt{3}}{\sqrt{3}}}$

$:\implies \bf{x = 100}$

$\boxed{\therefore \bf{x = 100\:m}}$

Therefore,

• The height of the tree , h = 100√3.

• The width of the river , x = 100 m.

Answer 2

Solution

•Let BC be the height of the tree,

•AB be the breadth of the river

•A be the initial position of the man.

•D be the final potion of the man angle CAB = 60°

and angle CDB = 30°

DA = 40 m

Let AB = x

Let BC = h

Consider the triangle DBC

$\sf \tan(30) = \dfrac{bc}{db} = \dfrac{bc}{da + ab} = \dfrac{h}{40 \times x}$

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

Using (2) in (1) we have

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

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

$\\ \sf \implies \: 3x = 40 + x$

$\\ \sf \implies \: 2x = 40$

$\\ \sf \implies \: x = 20$