Question

A tower stands vertically on the ground. From a point which is 15 meter away from the
foot of the tower, the angle of elevation of the top of the tower is 45°. What is the he
of the tower?
$\sqrt{123 \times 2233 \times 322 \times .2.. \div \times \times .1 = .026126101266}$
​

Answer 1

Answer:

Let the height of the tower = h m

Distance of the point of observation from the foot of the tower =15m

Angle of elevation of the top of the tower =45

∘

From the figure

tanθ=

adjacentsidetotheangleθ

oppositesidetotheangleθ

tan45

∘

=

15

h

⇒1=

15

h

⇒h=15×1=15

∴ The height of the tower =15m

Answer 2

Answer:

Let the height of the tower = h m

Distance of the point of observation from the foot of the tower =15m

Angle of elevation of the top of the tower =45  

∘

 

From the figure

tanθ=  

adjacentsidetotheangleθ

oppositesidetotheangleθ

​  

 

tan45  

∘

=  

15

h

​  

⇒1=  

15

h

​  

⇒h=15×1=15

∴ The height of the tower =15m