Question

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer 1

Answer:

The height of the tree is 8√3 m.

Step-by-step explanation:

Given :  

Let the broken part of the tree be AC & height of the remaining tree (AB)  be h m .

Distance between the foot of the tree to the point C is 8 cm i.e BC = 8 m

In right angle ∆ABC,

tan 30° =  P/B = AB/BC  

1/√3 = h/8

h = 8/√3 ………..(1)

In right angle ∆ABC,

sin 30° = P/H = AB /AC

½ = h / AC

½ = (8/√3)/AC

[From eq 1]

AC = 2 × 8/√3

AC = 16/√3 m

Height of the tree, H  = AB + AC  

H = 8/√3 + 16√3

H = (8 + 16)/√3

H = 24/√3

H = 24√3 / (√3 × √3)

[On Rationalising]

H = 24√3/3

H = 8√3 m

Hence, the height of the tree is 8√3 m.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

SOLUTION

Let AC was the original tree. Due to storm, it was broken into two parts.

The broken part A'B is making 30° with the ground.

In ∆A'BC,

$\frac{BC}{A'C} = tan30 \degree \\ \\ = > \frac{BC}{8} = \frac{1}{ \sqrt{3} } \\ \\ = > BC = (\frac{8}{ \sqrt{3} } )m \\ \\ \\ \frac{A'C}{A'B} = cos30 \degree \\ \\ = > \frac{8}{A'B} = \frac{ \sqrt{3} }{2} \\ \\ = > A'B = ( \frac{16}{ \sqrt{3} } )m$

Height of tree= A'B+ BC

$= > ( \frac{16}{ \sqrt{3} } + \frac{8}{ \sqrt{3} } )m = \frac{24}{ \sqrt{3} } m \\ \\ = > 8 \sqrt{3} m$

Hence, the height of the tree is 8√3m

hope it helps ☺️