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An observer 1.7m tall, is 20√3 m away from a tower. The angle of elevation from
the eye of the observer to the top of the to

By Isabella

An observer 1.7m tall, is 20√3 m away from a tower. The angle of elevation from
the eye of the observer to the top of the tower is 300
. Find the height of the tower

About the author
Isabella

Paheli is taking preparation for her final examination. She is writing some statements about
‘Nylon’. But she confused

multiply by using distributive property 12
[tex] \times [/tex]
25​

1 thought on “An observer 1.7m tall, is 20√3 m away from a tower. The angle of elevation from<br /> the eye of the observer to the top of the to”

  1. Step-by-step explanation:

    Let AB be the height of the tower and DE be the height of the observer.

    Then,

    In Δ ACD ,

    AC/DC = tan 30°

    ⇒ x/√30 = tan 30° = 1/√3

    ⇒ x = 20 m

    AB = 20 m + 1.7 m

    = 21.7 m

    So, the height of the tower is 21.7 m

    Reply

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