Question

1.5 m tall boy is standing at some distance from a 30 m tall building The ang
elevation from his eyes to the top of the building increases from 30 to 60°
towards the building. Find the distance he walked towards the building​

Answer 1

Answer:

Let the boy was standing at point S initially. He walked towards the building and reached at point T.

It can be observed that,

PR=PQ−RQ

⇒ PR=(30−1.5)m=28.5m=

2

57

In △PAR,

AR

PR

=tan30

o

⇒

2AR

57

=

3

1

∴ AR=

2

57

3

m

In △PRB,

BR

PR

=tan60

o

2BR

57

=

3

∴ BR=

2

3

57

=

2

19

3

m

We know, ST=AB

⇒ AB=AR−BR=(

2

57

3

−

2

19

3

)m

∴ AB=

2

38

3

m=19

3

m

solution

Step-by-step explanation:

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