Question

Two boats are sailing in the sea on either side of a light house. At a particular point of time, the angles of depression of the two boats as observed from the top of the light house are 45° and 30° respectively. If the light house is 100 m high, find the distance between the two boats. [HOTS] [Ans. 73.205 ml]​

Answer 1

Answer:

Let A and B be the position of two boats (figure)

Let PM be the lighthouse such that PM =100m

Let AM =x and BM =y

In ΔAPM,

PM

AM

=cot30

∘

⇒

100

x

=

3

⇒x=100

3

m

Similarly in Δ BPM,

PM

BM

=cot45

∘

⇒

100

y

=1⇒y=100m

∴ Required distance, AB=x+y

=100

3

+100=100(

3

+1)m=273.2m

Answer 2

Solution :

The distance between the two boats is 273.2 m.

Step by step explanation :

Let the position of first boat be P and second boat position be Q.

And let XY be the lighthouse.

Height of lighthouse ,XY=100 m

Let , PY= a and QY = b

We have to find ,The distance between the two boats i.e PQ

From the figure :

In ∆PYX

$\sf\tan30\degree=\dfrac{XY}{PY}$

$\sf\tan30\degree=\dfrac{100}{a}$

$\sf\dfrac{1}{\sqrt{3}}=\dfrac{100}{a}$

$\sf\:a=100\sqrt{3}$

Now , In ∆XYQ

$\sf\tan45\degree=\dfrac{XY}{QY}$

$\sf\tan45\degree=\dfrac{100}{b}$

$\sf1=\dfrac{100}{b}$

$\sf\:b=100$

Hence ,The distance between the two boats = PY+QY

= a+b

=100√3+100

=100(√3+1)

= 273.2 m

Therefore ,The distance between the two boats is 273.2 m.