Question

Two ships are approaching a light-house from opposite directions. The angles of depression of the two ships from the top of the light-house are 30o and 45o. If the distance between the two ships is 100 m, find the height of the light-house. [Use 3 = 1.732]

Answer 1

Please refer to the given diagram.
AB = 100
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

Answer 2

Answer:

Length of tower is 36.60 m

Step-by-step explanation:

Refer the attached figure

AB is the height of tower

The distance between the ships i.e. CD = 100 m

Let BD be x

So, CB = 100-x

The angles of depression of the two ships from the top of the light-house are 30° and 45° i.e. ∠ACB = 30° and ∠ADB = 45°

In ΔABD

$tan \theta =\frac{Perpendicular}{Base}$

$tan 45^{\circ}=\frac{AB}{BD}$

$1=\frac{AB}{x}$

$x=AB$   ---1

In ΔABC

$tan \theta =\frac{Perpendicular}{Base}$

$tan 30^{\circ}=\frac{AB}{BC}$

$\frac{1}{\sqrt{3}}=\frac{AB}{100-x}$

$\frac{100-x}{\sqrt{3}}=AB$  ---2

With 1 and 2

$x=\frac{100-x}{\sqrt{3}}$  

$\sqrt{3}x=100-x$  

$\sqrt{3}x+x=100$  

$(\sqrt{3}+1)x=100$  

$x=\frac{100}{(\sqrt{3}+1)}$  

$x=36.60$  

So, Length of tower is 36.60 m