Question

$\huge{\boxed{\underline{\bf{\red{Question:-}}}}}$
❐ A man standing on the deck of a ship, which is 10m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill.

Answer with proper explanation.
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Answer 1

Let a man is standing on the Deck of a ship at point a such that AB = 10 m & let CE be the hill

Thus, AB = CD = 10 m

The top and bottom of a hill is E and C.

Given, the angle of depression of the base C of the hill observed from A is 30° and angle of elevation of the top of the hill observed from A is 60 °

Then ∠EAD= 60° &

∠CAE= ∠BCA= 30°. (Alternate ANGLES)

Let AD = BC = x m & DE= h m

In ∆ ADE

tan 60° = Perpendicular / base = DE/AD

√3= h/x [tan 60° = √3]

h = √3x……..(1)

In ∆ ABC

tan 30° = AB /BC

[ tan30° = 1/√3]

1/√3 = 10/x

x= 10√3 m.. …………..(2)

Substitute the value of x from equation (2) in equation (1), we have

h = √3x

h= √3× 10√3= 10 × 3= 30 m

h = 30 m

The height of the hill is CE= CD+ DE= 10 +30= 40 m

Hence, the height of the hill is 40 m & the Distance of the hill from the ship is 10√3 m

Answer 2

Diagram

• According to diagram the ship is at point A and the total height of the hill is (x+10)m

Given :-

• Angle BAC = 60°
• Angle ADE = 30°
• CD = AE = 10 m

To Find :-

• Distance of hill from Ship(AC or DE)

• Height of the Hill ( BD )

Solution :-

• For finding the distance of ship from the hill and height of hill we will use the trigonometric ratio,

Take Triangle AED to find distance of Ship from the hill

$\sf\mapsto{\tan( \theta )=\dfrac{Perpendicular}{Base}}$

$\sf\mapsto{\tan(30)=\dfrac{AE}{DE}}$

$\sf\mapsto{\tan(30)=\dfrac{10}{DE}}$

$\sf\mapsto{\dfrac{1}{\sqrt{3}}=\dfrac{10}{DE}}$

$\sf{\boxed{\boxed{\red{\dag{DE=AC=10\sqrt{3}m}}}}}$

So , the distance of ship from heel hill is 10√3 m

• Take Triangle ABC to find the value of x

$\sf\mapsto{\tan(\theta)=\dfrac{Perpendicular}{Base}}$

$\sf\mapsto{\tan(60)=\dfrac{BC}{CA}}$

$\sf\mapsto{\tan(60)=\dfrac{x}{10\sqrt{3}}}$

$\sf\mapsto{\sqrt{3}=\dfrac{x}{10\sqrt{3}}}$

$\sf\mapsto{x= BC = 30m}$

Total height of Hill

$\implies\sf{BD = BC+CD}$

$\implies\sf{BD = 30+10}$

$\sf{\boxed{\boxed{\red{\dag{BD=40m}}}}}$

So , height of the hill is 40m