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Answers
Given :-----
- AB = Building = 30m
- angle AED = 30°
- angle AFD = 60°
- EC = Boy Height = 1.5m .
- FE = x metre , he walks Towards building .
[ All these values i took according to my diagram] .
Formula used :---
- Tan@ = Perpendicular /Base
- Tan30° = 1/√3
- Tan60° = √3 .
Solution. :-------
From image we can see that,
AB = 30m
DB = EC = 1.5m
Hence, AD = AB-BD = 30-1.5 = 28.5m.
Now, In ∆ADE, we have,
→ Tan30° = AD/DE
→ 1/√3 = 28.5/DE
→ DE = 28.5√3 m --------- Equation (1)
___________________________
Now, in ∆ADF we have,
→ Tan60° = AD/DF
→ √3 = 28.5/DF
→ DF = 28.5/√3
Rationalizing the RHS part we get,
→ DF = 28.5/√3 * (√3/√3) = 28.5√3/3 = 9.5m ----- Equation(2)
____________________________
Now, we have to Find FE = x metre .
→ FE = DE - DF
Putting values From both Equations we get,
→ X = 28.5√3 - 9.5√3 = 19√3m ..
______________________________
Hence, the boy walks 19√3 m Towards the building..
#BAL
#answerwithquality
Answer:
Given :-----
AB = Building = 30m
angle AED = 30°
angle AFD = 60°
EC = Boy Height = 1.5m .
FE = x metre , he walks Towards building .
[ All these values i took according to my diagram] .
Formula used :---
Tan@ = Perpendicular /Base
Tan30° = 1/√3
Tan60° = √3 .
Solution. :-------
From image we can see that,
AB = 30m
DB = EC = 1.5m
Hence, AD = AB-BD = 30-1.5 = 28.5m.
Now, In ∆ADE, we have,
→ Tan30° = AD/DE
→ 1/√3 = 28.5/DE
→ DE = 28.5√3 m --------- Equation (1)
___________________________
Now, in ∆ADF we have,
→ Tan60° = AD/DF
→ √3 = 28.5/DF
→ DF = 28.5/√3
Rationalizing the RHS part we get,
→ DF = 28.5/√3 * (√3/√3) = 28.5√3/3 = 9.5m ----- Equation(2)
____________________________
Now, we have to Find FE = x metre .
→ FE = DE - DF
Putting values From both Equations we get,
→ X = 28.5√3 - 9.5√3 = 19√3m ..
______________________________
Hence, the boy walks 19√3 m Towards the building..
#BAL
#answerwithquality