Question

ऑन द टॉप ऑफ द लाइट हाउस एंड ऑफ द वर्किंग एंड शीप मेक एंड एंगल ऑफ़ डिप्रेशन ऑफ 60 डिग्री​

Answer 1

Step-by-step explanation:

Let A and B be the two positions of the ship. Let d be the distance travelled by the ship during the period of observation i.e. AB = d metres.

Let the observer be at O, the top of the lighthouse PO.

It is given that PO = 100 m and the angle of depression from O of A and B are 30

∘

and 45

∘

respectively.

∴∠OAP=30

∘

and∠OBP=45

∘

In Δ OPB, we have

tan45

∘

=

BP

OP

⇒1=

BP

100

⇒ BP = 100 m

In Δ OPA, we have

⇒tan30

∘

=

AP

OP

⇒

3

1

=

d+BP

100

⇒ d + BP = 100

3

⇒ d + BP = 100

3

⇒ d = 100

3

- 100

⇒ d = 100(

3

- 1) = 100(1.732 - 1) = 73.2 m.

Hence, the distance travelled by the ship from A to B is 73.2 m.

Please follow me

Please mark me as Brainlist