Question

The upper part of a tree broken over by the wind makes an angle of 30° with the ground and the distance of the root from the point where the top touches the ground is 25 m. What was the height of the tree?

Answer 1

SOLUTION IS IN THE ATTACHMENT
LINE OF SIGHT: The line of sight is a line drawn from the eye of an observer to the point in the object viewed by the observer.

ANGLE OF ELEVATION: The angle of elevation of an object viewed is the angle formed by the line of sight with the horizontal , when it is above the horizontal level.

ANGLE OF DEPRESSION:The angle of depression of an object viewed is the angle formed by the line of sight with the horizontal , when it is below the horizontal level.

•Angle of elevation and depression are always acute angles.

•If the observer moves towards the perpendicular line(Tower/ building) then angle of elevation increases and if the observer move away from the perpendicular line(Tower/ building) angle of elevation decreases.

HOPE THIS WILL HELP YOU...

Answer 2

Hey !!

Here is your answer,,


The tree is of ( x+h ) height.

The (x) part of tree is broken and form 30° at distance of 25m away from its foot.

AB = h m
BC = 25m
AC = x m

Height of tree (x+h) = AB + AC


Tan 30 = P/b

1/√3 = AB/AC

1√3 = h/25

h = 25/√3

h = 14.43


Now, we have to find AC (x)...

AC^2 = AB^2 + BC^2

= (14.43)^2 + (25)^2

= 208.22 + 625

= 833.22

AC = √833.22

AC = 28.86m

Height of the tree (x+h) = AB + AC

= 14.43 + 28.86

= 43.29 m

Height of the tree is 43.29m


Hope it helps...

Thanks...