a 1.5 m tall boy is standing at some distance from of 39 M tall building the angle of elevation from his eyes to the top of building increases from 30° to 60° as you walk towards the building find the distance he walked toward the building?
Answers
Answer:
Given:-
Height of building (AB) = 39 m
height of man = 1.5 m
angle of elevation = 30° and 60°
A
|\.\
|...\..\
|.....\... \
|.......\....\
|.........\....\
|...........\.....\
|.60°b....\30°\
|_____(_\_ (\_D
B .............C
We have to find the value of CD.
In ∆ ABD
Tan 30°= AB
...............====
................BD
Hence ,
Tan 30° = 39
................====
.................BD
1...........39
== => ====
√3.......BD
=> BD= 39√3
Now
In ∆ABC
Tan 60°=.. AB
..................===
...............…BC
√3= 39
......====
........BC
BC√3 = 39
BC= 39
........===
........√3
BC= 39 .....√3
.......=== × ====
.......√3 .....√3
BC= 39√3
........=====
...........3
Now
CD= BD-BC
......................39√3
CD= 39√3 - =====
...........................3
..............117√3-39√3
CD = > ===========
.......................3
... ...........78√3
CD => ======
.................3
CD => 18√3
*Refer the attachment for figure.
Let BC = GF = y
CD = FE = x
Given : DE = 1.5 m, AG = 39 m
AB = AG - BG = 39 - 1.5 = 28.5 m
= 60° and = 30°
Find : Distance travelled by man towards building.
Solution :
In ∆ABD
=> tan30° =
=> =
=> =
Cross-multiply them
=> x + y = 28.5√3 _______ (eq 1)
___________________________
In ∆ABC
=> tan60° =
=> √3 =
=> y = m
Put value of y in (eq 1)
=> x + = 28.5√3
=> x = 28.5√3 -
=> x =
=> x =
=> x =
After rationalizing we get,
=> x = 19√3 m
____________________________