Question

What is the answer L















BRAINLIEST FOR. BRAINLIEST ​

Answer 1

Answer:

Step-by-step explanation:

Let the height of the building = AB = OC = 50 m. And

height of the tower = DC = ( DO + OC ) = x m.

Distance between tower and building = AC = BO

Angle of depression to the top of building =∠EDB = 45°

Angle of depression to bottom of building = ∠EDA = 60°

In right angle ΔDBO,

tan θ = DO / BO

tan 45° = x / BO

1 = x / BO

BO = x m.

In right angle ΔDAC,

tanθ = DC / AC

tan 60° = ( DO + OC ) / AC

√3 = ( x + 50 )/ AC

AC = ( x + 50 ) / √3

We know that BO = AC

∴ x = (x + 50) / √3

⇒√3x = x + 50

⇒√3x - x = 50

⇒ x ( √3 - 1 ) = 50

⇒ x = 50 / (√3- 1 )

= 50 (√3+ 1 ) / (√3 - 1 ) ( √3 + 1 ) (∵Rationalising the denominator )

= 50( √3 + 1 ) / ( √3)² - ( 1 )²

= 50 (√3 + 1 ) / 3 - 1

= 50 (√3 + 1 ) / 2

= 25 (√3 + 1 )

= 25√3 + 25

= 25 ( 1.73 ) + 25

= 43.25 + 25

= 68.25 m.

∴BO = AC = x = 68.25 m.

Hence, distance between building and tower= AC = 68.25 m.

∴Height of the tower = DC = DO + AC = 68.25 m + ( x + 50 / √3 ) m.

= 68.25 m + ( 68.25 + 50 / √3 ) m

= 68.25 m + ( 118 / √3 ) m

= 68.25 m + ( 118 (√3 ) / (√3)² )

= 68.25 m + ( 118 × 1.73 / 3 )

= 68.25 m + ( 204.14 / 3)

= 68.25 m + 68.04 m

= 136.29 m.

Answer 2

Answer:

Bro this is long but cool to slolve (AND MARK BRAINLIEST)

Step-by-step explanation:

Let the height of the building = AB = OC = 50 m. And

height of the tower = DC = ( DO + OC ) = x m.

Distance between tower and building = AC = BO

Angle of depression to the top of building =∠EDB = 45°

Angle of depression to bottom of building = ∠EDA = 60°

In right angle ΔDBO,

tan θ = DO / BO

tan 45° = x / BO

1 = x / BO

BO = x m.

In right angle ΔDAC,

tanθ = DC / AC

tan 60° = ( DO + OC ) / AC

√3 = ( x + 50 )/ AC

AC = ( x + 50 ) / √3

We know that BO = AC

∴ x = (x + 50) / √3

⇒√3x = x + 50

⇒√3x - x = 50

⇒ x ( √3 - 1 ) = 50

⇒ x = 50 / (√3- 1 )

= 50 (√3+ 1 ) / (√3 - 1 ) ( √3 + 1 ) (∵Rationalising the denominator )

= 50( √3 + 1 ) / ( √3)² - ( 1 )²

= 50 (√3 + 1 ) / 3 - 1

= 50 (√3 + 1 ) / 2

= 25 (√3 + 1 )

= 25√3 + 25

= 25 ( 1.73 ) + 25

= 43.25 + 25

= 68.25 m.

∴BO = AC = x = 68.25 m.

Hence, distance between building and tower= AC = 68.25 m.

∴Height of the tower = DC = DO + AC = 68.25 m + ( x + 50 / √3 ) m.

= 68.25 m + ( 68.25 + 50 / √3 ) m

= 68.25 m + ( 118 / √3 ) m

= 68.25 m + ( 118 (√3 ) / (√3)² )

= 68.25 m + ( 118 × 1.73 / 3 )

= 68.25 m + ( 204.14 / 3)

= 68.25 m + 68.04 m

= 136.29 m.

Let the height of the building = AB = OC = 50 m. And

height of the tower = DC = ( DO + OC ) = x m.

Distance between tower and building = AC = BO

Angle of depression to the top of building =∠EDB = 45°

Angle of depression to bottom of building = ∠EDA = 60°

In right angle ΔDBO,

tan θ = DO / BO

tan 45° = x / BO

1 = x / BO

BO = x m.

In right angle ΔDAC,

tanθ = DC / AC

tan 60° = ( DO + OC ) / AC

√3 = ( x + 50 )/ AC

AC = ( x + 50 ) / √3

We know that BO = AC

∴ x = (x + 50) / √3

⇒√3x = x + 50

⇒√3x - x = 50

⇒ x ( √3 - 1 ) = 50

⇒ x = 50 / (√3- 1 )

= 50 (√3+ 1 ) / (√3 - 1 ) ( √3 + 1 ) (∵Rationalising the denominator )

= 50( √3 + 1 ) / ( √3)² - ( 1 )²

= 50 (√3 + 1 ) / 3 - 1

= 50 (√3 + 1 ) / 2

= 25 (√3 + 1 )

= 25√3 + 25

= 25 ( 1.73 ) + 25

= 43.25 + 25

= 68.25 m.

∴BO = AC = x = 68.25 m.

Hence, distance between building and tower= AC = 68.25 m.

∴Height of the tower = DC = DO + AC = 68.25 m + ( x + 50 / √3 ) m.

= 68.25 m + ( 68.25 + 50 / √3 ) m

= 68.25 m + ( 118 / √3 ) m

= 68.25 m + ( 118 (√3 ) / (√3)² )

= 68.25 m + ( 118 × 1.73 / 3 )

= 68.25 m + ( 204.14 / 3)

= 68.25 m + 68.04 m

= 136.29 m.