❐ A man standing on the deck of a ship, which is 10m above water le

1.	❐ 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.Kindly don't spam!
Answer» 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