Question

3. As observed from the top of a light house, 100 m above sea level, the angle of depression of a
ship, sailing directly towards it, changes from 30° to 60°. Determine the distance travelled by the
ship during the period of observation. [Take V3 = 1.732]
and the same side of it are​

Answer 1

Answer:

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.

solution

Step-by-step explanation: