Question

. From the top and foot of a tower 40 m high, the angle of elevation of the top of a lighthouse
is found to be 30° and 60° respectively. Find the height of the lighthouse. Also find the
distance of the top of the lighthouse from the foot of the tower.
​

Answer 1

Solution :-

Consider the tower be AC and height of the tower AC is 40m high .

Let take the base of the tower AB be xm

Therefore, CD is also xm

Angle of elevation ΔDCE = 30°

Angle of depression ΔBAE = 60°

Let the DE be h

Therefore,

In a right angled triangle CDE,

Tan30° = DE /CD

Tan30° = h / x

1 / √3 = h / x

x = √3h. eq( 1 )

Now, In right angled triangle ABE

Tan 60° = BE / AB

√3 = h + 40 / x

x = h + 40 / √3. eq( 2.)

Comparing eq( 1 ) and eq( 2 )

√3h = h + 40/√3

√3h * √3 = h + 40

3h = h + 40

3h - h = 40

2h = 40

h = 40/2

h = 20m

Thus, Height of the light house = 40 + 20 = 60m

Now,

In right angled triangle ABE

Sin 60° = BE / AE

√3/2 = 60/AE

AE = 60 * 2 / √3

AE = 120 / √3

By rationalisation

AE = 120/√3 * √3/√3

AE = 120√3/3

AE = 40√3 m $.$

Answer 2

Answer:

$\huge \underline \mathfrak \red{Solution}$

Consider the tower be AC and height of the tower AC is 40m high .

Let take the base of the tower AB be xm

Therefore,

• CD is also xm

Angle of elevation ΔDCE = 30°

Angle of depression ΔBAE = 60°

Let the DE be h

Therefore,

In a right angled triangle CDE,

Tan30° = DE /CD

Tan30° = h / x

1 / √3 = h / x

x = √3h. eq( 1 )

Now, In right angled triangle ABE

Tan 60° = BE / AB

√3 = h + 40 / x

x = h + 40 / √3. eq( 2.)

Comparing eq( 1 ) and eq( 2 )

√3h = h + 40/√3

√3h * √3 = h + 40

3h = h + 40

3h - h = 40

2h = 40

h = 40/2

h = 20m

Thus, Height of the light house = 40 + 20 = 60m

Now,

In right angled triangle ABE

Sin 60° = BE / AE

√3/2 = 60/AE

AE = 60 * 2 / √3

AE = 120 / √3

By rationalisation

AE = 120/√3 * √3/√3

AE = 120√3/3

AE = 40√3 m