As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are are 30° and 45°.if one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Answers
Given :
- As observed from the top of a 75 m high lighthouse from the sea-level.
- The angles of depression of two ships are are 30° and 45°.
- One ship is exactly behind the other on the same side of the lighthouse.
To find :
- The distance between the two ships =?
Step-by-step explanation:
Refer to the attached pic.
B and C are the two ships and they are separated by the distance BC .
Let BC be x .
The height of the lighthouse is given 75 m .
Hence AD= 75 .
Use √3 as 1.732 .
Also use tan Ф = ( side opposite to Ф ) / ( side adjacent to Ф )
Use tan 45 = 1 .
Use tan 30 = 1/√3 .
∠ACB = 30°
∠ADB = 45°
In Δ ABD ,
tan 45° = AB /BD
➮ tan 45° = 75/BD
➮ 1 = 75/BD
➮ BD = 75
From the figure we have BD + DC = BC
➮ 75 + x = BC
In Δ ABC ,
tan 30° = AB/BC
➮ 1/√3 = 75 / ( 75 + x )
➮ 75 + x = 75√3
➮ x = 75√3 - 75
➮ x = 75 ( √3 - 1 )
➮ x = 75 ( 1.732 - 1 )
➮ x = 75 × 0.732
➮ x = 54.9
Therefore, The distance between the ships is 54.9 m .
Solution:
◕Refer the attachment
✧Let DC be the light house of 75 from the sea level the angle of depression of one ship is 30° and other ship B is 45° .Such that DAC = 30° and DBC = 45° .We need to find the distance between two ships i.e. A and B (distance).
◕In ∆ DBC,
→ DC/BC = tan45°
→75/BC = 1
→BC = 75 m
In ∆ DAC,
→DC/AB+BC = tan30°
*suppose AB = x
→ 75/(x+75) = 1/√3
→x+75 = 75√3
→x = 75√3-75
→x= 75(√3-1)m
Hence the distance between two ships A and = 75(√3-1)m
