Question

As observed from the top of a 100 metre high lighthouse from the sea level the angles of depression of two ships are 30 degree and 45 degree if one ship is exactly behind the other on the same side of the lighthouse find the distance between the two ships

Answer 1

AB-100m
angle D is 45° and angle C is 30°
so,ABD & ABC is a right angle triangle.
therefore,
∆ABD
AB/BD=tan45°
or,100/BD=1
or,BD=100m

∆ABC
AB/BC=tan30°
or,100/100+x=1/√3. DC=x,BC=BD+CD
or,100√3=100+x
or,100√3-100=x
or,100(√3-1)=x
so...DC is 100(√3-1)m
The distance between the two ship is 100(√3-1)m

Answer 2

BEFORE ANSWERING TO THIS QUESTION LET ME REMIND YOU THAT THIS QUESTION WAS ASKED IN THE 10TH CBSE BOARD EXAMINATION 2017 2018 AND I HAVE WRITTEN IT!!

SO COMING TO THE QUESTION,

IN THE FIGURE,

+AB is the height of the lighthouse

+C and D are the positions of the two ships respectively.

+BC is the the distance between the first ship and the Foot of the Lighthouse

+BD is the distance between the second ship and the foot of the Lighthouse

GIVEN, AB=100m

TO FIND: CD= x

$\tan(30) = 1 \div \sqrt{3}$

>AB/BD=1/√3
>100/BD=1/√3
>100√3=BD (1)

$\tan(45) = 1$

>AB/BC=1
>100/BC=1
>BC=100 (2)


FROM (1),
>BD=100√3
>BC+CD=100√3

FROM (2),
>100+x=100√3
>x=100√3-100
>x=100(√3-1)
>x=100(1.73-1)
>x=100X 0.73
>x=CD=73m