Question

Angle of elevation is the angle which line of sight makes with the horizontal. Angle of elevation of the top of a tall building is 30° from a place A and becomes 60° from another place B that is 10√3 m from A towards the building as
shown in the figure. Height of the building is close to
​​

Answer 1

Answer:

Height of the building is 15 m

Step-by-step explanation:

let the height of building be x

distance from building to B be y

Answer 2

Given :-

• Angle of elevation from A = 30°
• Angle of elevation from B = 60°
• The horizontal base from A to B = 10√3m

To Find :-

• Height of building close to A or B = ?

Solution:-

• To calculate the height of building at first we have to assume the height building be h and the horizontal base from A to D be x . As per the given figure in ∆ ACD angle of elevation with 30 ° and In ∆ BCD angle of elevation with 60°.

Calculation begin :-

• From ∆ ACD

In right angle triangle ACD :-

➞ Tan 30° = CD/AD

• Here CD = Height of building (h)
• Here AD = Total base (AB + BD)
• Here AB = 10√3 m. BD = x

➞ 1/√3 = h/10√3 + x

➞ 10√3 + x = h√3

➞ h = (10√3 + x)/√3 ---------(i)

• From ∆ BCD

In right angle triangle BCD :-

➞ Tan 60° = CD/BD

➞ √3 = h/x

➞ h = x√3 ---------(ii)

• Putting together EQ (I) and (ii):-

➞ h = x√3

➞ (10√3 + x)/√3 = x√3

➞ 10√3/√3 + x/√3 = x√3

➞ 10 + x/√3 = x√3

➞ x√3 - x/√3 = 10

➞ (x√9 - x)/√3 = 10

➞ (3x - x)/√3 = 10

➞ 2x = 10√3

➞ x = 5√3 m

• Now calculate the height of building:-

➞ h = x√3 (from eq (ii))

➞ h = (5√3) × √3

➞ h = 5×√9

➞ h = 5 × 3

➞ h = 15m

Therefore,

• The height of building = 15 m

By calculating we get on conclusion that The height of building is close to B . As per above calculation from a point A to B is total distance AB = 10√3 but from a point B to D is total distance BD = 5√3 m.