Question

An observer 2 m tall is 10√3 m away from a tower. the angle of elevation from his eye to the top of the tower is 30º. the height of the tower is:

Answer 1

Heya !!!



Let AB be the tower and CD be the observer.



Draw DE perpendicular AB


Then AE = CD = 2 m


Let AB = X m .



Then ,


BE = ( X - 2 ) m


and



Angle EDB = 30°



Also,



DE = CA = 10 root 3 m

From right angle ∆ BED , we have



BE / DE = Tan 30°




=> ( X - 2 ) / 10 root 3 = 1/ root 3






=> ( X -2) = 1 × 10 root 3 / root 3






=> X -2 = 10



=> X = 10 +2



=> X = 12 m




Hence,



Height of tower is 12 m




★ ★ ★ HOPE IT WILL HELP YOU ★ ★ ★

Answer 2

Answer: Hoy!!

Height of the man = 2m

Angle of elevation = 30°

Distance from ground = 10√3 m

x/10√3 = 1/√3

x = 10√3/√3

x = 10m

ADD 2m to X,

So, x + 2 = 10 + 2 = 12m.

Thnku ya!!!