Question

iv) Two buildings are facing each other on a road of width 15 metres. From the top
of the first building, having a height of 12 metres, the angle of elevation of the
top of the second building is 30°. What is the height of the second building?​

Answer 1

$\textbf{Given that ,}$

□ Two buildings are in font of each other on a road of width ( base ) = 15 m.

□ Height of the first building ( perpendicular ) = 12 m

□ And the angle of elevation of the top of the second building from first building = 30°.

□ Height of the second building = ?

□ Let's solve this question ,

______________________________________________________________

□ Let's suppose the height of the another remaining building be " x "

□ Then height of second building = " x " + height of first building.

□ Height of second building = x + 12.

□ Base of the building = base at the top of building.

=> 15 = 15

According to the question ,

=> Tan30° = perpendicular / base

[ Value of Tan30° = 1 / √3 ]

=> 1 / √3 = x / 15

=> x√3 = 15

=> x = 15 / √3

□ Now , we have to multiply 15 /√3 by √3 / √3

=> x = ( 15 /√3 ) × (√3 / √3 )

=> x = ( 15√3 ) / 3

=> x = 5√3

Hence , height of the remaining building = 5√3

[ Value of √3 = 1.73 ]

Now ,

=> 5 × 1.73

=> 8.65

□ Now , total height of the building = x + height of the first building [ as both are parallel ].

=> 8.65 + 15 = 23.65

□ Hence , the total height of the another building is 23.65 m. or 5√3 + 12.

$\textbf{Thanks !!}$

$\textbf{be brainly}$

Answer 2

$\large \red {Given\:that}$

□ Two buildings are in font of each other on a road of width ( base ) = 15 m.

□ Height of the first building ( perpendicular ) = 12 m

□ And the angle of elevation of the top of the second building from first building = 30°.

□ Height of the second building = ?

□ Let's solve this question ,

______________________________________________________________

□ Let's suppose the height of the another remaining building be " x "

□ Then height of second building = " x " + height of first building.

□ Height of second building = x + 12.

□ Base of the building = base at the top of building.

=> 15 = 15

According to the question ,

=> Tan30° = perpendicular / base

[ Value of Tan30° = 1 / √3 ]

=> 1 / √3 = x / 15

=> x√3 = 15

=> x = 15 / √3

□ Now , we have to multiply 15 /√3 by √3 / √3

=> x = ( 15 /√3 ) × (√3 / √3 )

=> x = ( 15√3 ) / 3

=> x = 5√3

Hence , height of the remaining building = 5√3

[ Value of √3 = 1.73 ]

Now ,

=> 5 × 1.73

=> 8.65

□ Now , total height of the building = x + height of the first building [ as both are parallel ].

=> 8.65 + 15 = 23.65

□ Hence , the total height of the another building is 23.65 m. or 5√3 + 12.