an observer 1.7 M tall tall 20 root 3 M away from a tower the angle of elevation from the eye of the observer to the top of a tower is 30 degree find the height of the tower
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Solution :-
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
Answer.
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
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Hope it helps you!
Regards,
Shobana
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