The angle of elevation of the top of a tree from a point 50 feet from it's base is 30 degrees . What is the height of the tree.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The angle of elevation from a point 50 feet from the base of the tree to the top of the tree is 30 degrees.

Now tan 30 = height of tree/ distance of the point from the base

=> tan 30 = H / 50

=> 1/ sqrt 3 = H / 50

=> H = 50/sqrt 3

=> 28.86 feet

The height of the tree is 28.86 feet.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

LetĀ  B be the base of the tree, T the top of the tree. Let P be the point 50 ft from the base B.

So BP = 50 ft.

BT = height of the tree to be determined.

Angle BPT = 30 deg.

Angle PBT = 90 degree.

ThereforeĀ  in the right angled triangle, BT/BP = tan30.

Therefore BT = BP*tan30 deg

BT = 50* (1/sqrt3) ft.= 28.87 ft nearly.

Therefore the height of the tree is 28.87 ft.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We notice that it is formed a right triangle, whose base is of 50 feet long and it's horizontal angle is of 30 degrees.

We notice that the base is the adjacent side from the angle of 30 and the height is the opposite side from the angle.

We'll apply tangent function:

tan 30 = height/50

1/sqrt3 = height/50

height = 50/sqrt3

height = 50*sqrt3/3

The height of the tree is: height = 28.867 feet

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