Question

from a point on the ground the angles of elevation of the top and bottom of a transmission tower fixed at the top of a 20m high building are 60°&45°respectively find the height of the transmission tower​

Answer 1

$\sf\small\underline\red{Given:-}$

from a point on the ground the angles of elevation of the top and bottom of a transmission tower fixed at the top of a 20m high building are 60° & 45° respectively.

$\sf\small\underline\red{To \: Find:-}$

find the height of the transmission tower.

$\sf\small\underline\red{Solution:-}$

Let DC be the tower and BC be the building.

Then,

$\sf\small{∠CAB=45°}$

$\sf\small{∠DAB=60°}$

$\sf\small{BC=20 m}$

$\sf\small{Let \: height \: of \: the \: tower,}$

$\sf\small{DC=h \: m}$

$\sf\small{In \: right \: △ABC,}$

$\sf\small{tan45° = \frac{BC}{AB} }$

$\sf\small{⟶1 = \frac{20}{AB} }$

$\sf\small{⟶AB=20 \: m}$

$\sf\small{In \: right \: △ABD,}$

$\sf\small{tan60° = \frac{BD}{AB}}$

$\sf\small{⟶√3 = \frac{h + 20}{20} }$

$\sf\small{⟶h=20( √3 −1) m}$

Answer 2

Given :

from a point on the ground the angles of elevation of the top and bottom of a transmission tower fixed at the top of a 20m high building are 60° and 45° respectively.

To Find :

find the height of the transmission tower.

Solution :

Let DC be the tower and BC be the building.

then,

angle CAB = 45°

angle DAB = 60°

BC = 20m

Let the height of the tower ,

DC = h m

In right ΔABC,

tan 45° = BC/AB

➡️ 1 = 20/AB

➡️AB = 20 m

In right ΔABD,

tan60° = BD/AB

➡️ √3 = h + 20/20

➡️ h = 20(√3 – 1)m