Question

A person who is sitting on the ship of height 5m is observing the top of a submarine at 60° and bottom of the submarine at 30° . Find the distance between submarine and ship. Also find the height of the submarine and distance between top of submarine and the observer.

Please answer this question ☺️​

Answer 1

$\large{\underline{\bf{\pink{Answer:-}}}}$

✰ The height of the submarine = 5√3m

$\large{\underline{\bf{\blue{Explanation:-}}}}$

$\large{\underline{\bf{\green{Given:-}}}}$

✰ ship is at the height of 5m above water.

✰ angle of elevation to the top of the submarine = 60°

✰angle of depression to the bottom of submarine = 30°

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to find the height of the submarine and the distance between the ship and submarine.

$\huge{\underline{\bf{\red{Solution:-}}}}$

Let AB be the distance of ship from the water.

CD be the submarine.

Let man be at B point.

Then AB =5m

Let BE ⊥ CD and AC ⊥ CD

then,

$\angle\:EBD=60\degree$

and $\angle\:EBC=30\degree$

$\angle\:ACB=\angle\:EBC=30\degree[Alternate\: angles$

Let CD =h meters .

Then, CE = AB = 5m

and ED = (h - 5)m

In right angled triangle CAB:-

$\frac{AC}{AB}=cot30\degree$

$:\implies\purple{\:cot30\degree=\sqrt{3}}$

Then,

$:\implies\:\frac{AC}{5}=\sqrt{3}$

$:\implies\bf{\red{\:AC=5\sqrt{3}m}}$

So, BE=AC =5√3m

In right angled∆BED:-

$\frac{DE}{BE}=tan60\degree$

$:\implies\:\purple{tan60\degree=\sqrt{3}}$

Then,

$:\implies\:\frac{h-5}{5\sqrt{3}}=\sqrt{3}\:\:\:\:\:\:[using\:(i)]$

$:\implies\:h-5=5\times\:3$

$:\implies\:h-5=15$

$:\implies\:h=15+5$

$:\implies\bf{\red{\:h=20m}}$

hence,

The height of submarine = 20m

and the distance between submarine and ship= 5√3

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Answer 2

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{Distance \ between \ ship \ and }$

$\bold{submarine \ is \ 5\sqrt3 \ m \ and }$

$\bold{height \ of \ submarine \ is \ 20 \ m}$

$\bold\orange{Given:}$

$\bold{=>Ship \ is \ at \ height \ of \ 5 \ m}$

$\bold{=>Angle \ of \ elevation \ is \ of \ 60°}$

$\bold{=>Angle \ of \ depression \ is \ of \ 30°}$

$\bold\pink{To \ find:}$

$\bold{=>Distance \ between \ ship \ and \ submarine}$

$\bold{=>Height \ of \ submarine}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{Let \ AB \ represent \ the \ ship \ and \ CE}$

$\bold{represent \ the \ submarine.}$

___________________________________

$\bold{Angle \ of \ elevation=60°}$

$\bold{Therefore, \ Angle(EBD)=60°}$

$\bold{Angle \ of \ depression=30°}$

$\bold{Therefore, \ Angle(DBC)=30°}$

____________________________________

$\bold{Angle(BCA)=Angle(DBC)... Alternate \ angles}$

$\bold{Therefore, \ Angle(DCB)=30°}$

$\bold{In \ TriangleABC, \ Angle(BAC)=90°}$

$\bold{Angle(ABC)=60°... remaining \ angle}$

$\bold{BA=5 \ m...given}$

$\bold{tan30°=\frac{BA}{CA}}$

$\bold{But,tan30°=\frac{1}{\sqrt3}}$

$\bold{\frac{BA}{CA}=\frac{1}{\sqrt3}}$

$\bold{\frac{5}{CA}=\frac{1}{\sqrt3}}$

$\bold{CA=5\sqrt3}$

_____________________________________

$\bold{In \ quadrilateral (ABCD),}$

$\bold{Angle(ABC)=90°}$

$\bold{Angle(DCA)=90°}$

$\bold{Angle(ABD)=Angle(ABC)+Angle(CBD)}$

$\bold{Angle(ABD)=60°+30°=90°}$

$\bold{Angle (DCA)=90°... remaining \ angle}$

$\bold{Therefore, \ quadrilateral (ABCD) \ is \ a \ rectangle}$

$\bold{By \ property \ of \ rectangle,}$

$\bold{AB=CD=5 \ m}$

$\bold{AC=BD=5\sqrt3 \ m}$

$\bold{Therefore, \ distance \ between \ ship \ and }$

$\bold{submarine \ is \ 5\sqrt3 \ m}$

______________________________________

$\bold{In \ TriangeEBD,}$

$\bold{Angle (EBD)=60°...given}$

$\bold{Angle(EDB)=90°...linear \ angle \ axiom}$

$\bold{BD=5\sqrt3 \ m}$

$\bold{tan60°=\frac{ED}{BD}}$

$\bold{But,tan60°=\sqrt3}$

$\bold{Therefore,\frac{ED}{BD}=\sqrt3}$

$\bold{\frac{ED}{5\sqrt3}=\sqrt3}$

$\bold{ED=15 \ m}$

$\bold{Height \ of \ submarine=ED+DC}$

$\bold{Height \ of \ submarine=15+5}$

$\bold{Height \ of \ submarine=20 \ m}$

$\bold{Therefore,}$

$\bold\purple{Distance \ between \ ship \ and }$

$\bold\purple{submarine \ is \ 5\sqrt3 \ m \ and }$

$\bold\purple{height \ of \ submarine \ is \ 20 \ m}$