Question

Two ships are approaching a light house from opposite directions. The angles of depression of two ships from the top of light house are 30 degree and 45 degree .If the distance between two ships is 100 m ,find the height of light house

Answer 1

EXPLANATION.

→ AD be the light house and height = h

→ in ∆ADB

→ tan ø = perpendicular/base = p/b

→ tan (45°) = AD/BD

→ tan (45°) = h/x [ tan (45°) = 1 ]

→ 1 = h/x

→ h = x .......(1)

→ in ∆ADC

→ tan ( 30°) = AD/DC

→ tan (30°) = h / 100 - x

→ 1/√3 = h / 100 - x

→ 1/√3 = h / 100 - h [ h = x ]

→ 100 - h = √3h

→ 100 = √3h + h

→ 100 = h ( √3 + 1 )

→ 100/√3 + 1 = h

→ 100 / 1.732 + 1 = h [ √3 = 1.732 ]

→ 100 / 2.732 = h

→ h = 36.6 m

→ height of lighthouse = 36.6 m.

Answer 2

Step-by-step explanation:

Given : -

• Two ships are approaching a light house from opposite directions.

• The angles of depression of two ships from the top of light house are 30 degree and 45 degree .

• If the distance between two ships is 100 m

To Find : -

• find the height of light house

Solution : -

Let AC = x

Let BC = 100 - x

Let CD = h

In ΔEAD,

EA = DC

ED = AC

tan 45° = EA/ED

1 = h/x

x = h

In ΔDFB

FB = CD

DF = CB

tan 30° = FB/DF

1/√3 = h/100 - x

1/√3 = h/100 - h

100 - h = √3h

100 - h = 1.732h

2.732h = 100

h = 100/2.732

h = 36.603m

h = 36.6m

Hence the. answer is 36.6m