Question

The upper part of a tree, broken over by the wind, makes an angle of 45° with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. What was the height of the tree before it was broken?

Answer 1

From figure,

In ΔABC,

∴ `tan 45^circ = "Perpendicular"/"Base"`

1 = `(AB)/(BC)`

AB = BC = 15 meters

In right angle triangle ABC,

AC² = AB² + BC²

AC² = 15² + 15²

AC² = 225 + 225

AC² = 450

AC = `sqrt(450)`

= `15sqrt2`

Height of tree = AB + AC 

= `15 + 15sqrt2`

= 15 + 21.21

= 36.21 m

Hence, the height of the tree before it was broken = 36.21 metres.