Question

32. The shadow of a tower standing on a level ground is found to be 40 m longer when the
Sun's altitude is 45° than when it is 60°. Find the height of the tower.​

Answer 1

Answer:

In △ABD,

tan60

o

=

BD

AB

3

=

x

h

∴ x=

3

h

------- ( 1 )

Now in △ABC,

tan30

o

=

BC

AB

⇒

3

1

=

x+40

h

⇒ x+40=

3

h

∴

3

h

+40=

3

h [ From equation ( 1 ) ]

⇒ h+40

3

=3h

⇒ 2h=40

3

∴ h=20

3

m

∴ The height of a tower is 20

3

m.

solution