Question

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?​

Answer 1

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

Answer 2

*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

$\angle{ACB}$ = 60° and $\angle{ADB}$ = 30°

Find : Distance travelled by man towards building.

Solution :

In ∆ABD

=> tan30° = $\dfrac{AB}{BD}$

=> $\dfrac{1}{\sqrt{3}}$ = $\dfrac{AB}{BC\:+\:CD}$

=> $\dfrac{1}{\sqrt{3}}$ = $\dfrac{28.5}{x\:+\:y}$

Cross-multiply them

=> x + y = 28.5√3 _______ (eq 1)

___________________________

In ∆ABC

=> tan60° =$\dfrac{AB}{BC}$

=> √3 = $\dfrac{28.5}{y}$

=> y = $\dfrac{28.5}{\sqrt{3}}$ m

Put value of y in (eq 1)

=> x + $\dfrac{28.5}{\sqrt{3}}$ = 28.5√3

=> x = 28.5√3 - $\dfrac{28.5}{\sqrt{3}}$

=> x = $\dfrac{28.5(3)\:-\:28.5}{\sqrt{3}}$

=> x = $\dfrac{85.5\:-\:28.5}{\sqrt{3}}$

=> x = $\dfrac{57}{\sqrt{3}}$

After rationalizing we get,

=> x = 19√3 m

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