Question

27. From a point on the ground, the angles of elevation of the top and bottom of
transmission tower fixed at the top of a 20m high building are 60% and 45°
respectively. Find the height of the transmission tower

please help me
step by step explanation need in 15 minutes
​

Answer 1

Answer:

case 1 - in triangle DCB

angle c = 90°

and angle CDB = 45°

NOW, tan 45°= BC/CD

tan 45° = 20/CD

1 = 20/CD (tan 45° =1)

so CD = 20 m

now case 2

in triangle ACD angle A= 90°

and angle ADC = 60°

SO, tan 60° = AC/CD

√3 = AC /20 ( tan 60° = √3)

AC = 20√3

hence height of the transmission tower =

AB = AC -BC

AB = 20√3 - 20

AB = 20(√3 -1)

answer. . . . .

Answer 2

$\underline{\underline{\maltese\: \: \textbf{\textsf{ Question }}}}$

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

$\underline{\underline{\maltese\: \: \textbf{\textsf{Answer}}}}$

Let DC be the tower and BC be the building. Then,

∠CAB=45°

∠DAB=60°

BC=20 m

Let height of the tower,

DC = h m

In right △ABC,

$\sf \: tan \: 45° = \frac{BC}{AB} \\ \\ \sf \implies 1= \frac{20}{AB} \\ \\ \sf \implies AB=20 \: m$

In right △ABD,

$\sf \: tan60° = \frac{BD}{AB} \\ \\ \sf \implies \sqrt{3} = \frac{h+20}{20} \\ \\ \sf \implies h=20( \sqrt{3} −1) m$

Height of tower is 20(√3-1) m.