## How to find the height of a tower for unknown angles?

For better understanding take an example. Let's discuss-

*Q-1) The angle of elevation of the top of a tower from two points at a distance of 4m & 9m from the base of the tower & in the same straight line with it are complementary. Find the height of the tower.*

Solve:-

Let D & B are the points which are 4m & 9m away from the base of the tower (

*Point C*).According to the question, Angle making by point B & D to point A are complements to each other.

Sum of complementary angles is 90Âº.

Given: BC = 9m, DC = 4m

Let∠ABD = Î¸, So ∠ADC = (90Âº-Î¸)

In ∆ABC

tanÎ¸ = AC / BC

*{ tan = Perpendicular / Base }*tanÎ¸ = AC / 9 .......Eq.1

In ∆ADC

tan(90ÂºÎ¸) = AC / DC

cotÎ¸ = AC / 4

*{ cotÎ¸= tan(90Âº-Î¸) }*1/tanÎ¸ = AC / 4

*{ cotÎ¸ = 1/tanÎ¸ }*tanÎ¸ = 4 / AC ......Eq.2

On equating values of Eq.1 & Eq.2

AC/9 = 4/AC

AC² = 9×4

AC² = 36

AC = √36

AC = 6

So the height of the tower will be 6m.

## How to find the height of a tower if it is placed on the building?

Let's discuss

*Q-2) From a point on the ground, the angle of elevation of the bottom & top of a transmission tower fixed at the top of a 20m high building is 45Âº & 60Âº respectively. Find the height of the tower.*

Solve:-

Let DB = Height of building (20m)

AD = Height of Tower,

Point C makes an angle of 45Âº with D

*(Bottom of the tower)*.& Point C makes an angle of 60Âº with point A

*(Top of the tower)*.In ∆ABC

tan 60Âº = AB / BC

*{ Value of tan60Âº = √(3) }**√*3 = (AD + DB) / BC

*{ AB = AD + DB }*

√3 = (AD + 20) / BC .......Eq.1

In ∆DBC

tan45Âº = DB / BC

*{ Value of tan45Âº = 1 }*1 = 20 / BC

BC×1 = 20

BC = 20m ........Eq.2

Put the value of BC in Eq.1

√3 = (AD + 20) / 20

20√3 = AD + 20

20√3 - 20 = AD

20(√3 - 1) = AD

We get the height of tower 20(√3 -1)m.