Question

A tree breaks due to storm and the broken parts bend such that the top of the tree touches the ground making an angle 30 with it. The distance between the foot of the tree to the point where the top touches the ground is 10 m. Then the height of the tree is:​

Answer 1

Answer:

Let AD is the broken part of the tree.

So total length of the tree = AB + AD

Again AD = AC

So total length of the tree = AB + AC

Now in triangle ABC,

cos 30 = BC/AC

=> √3/2 = 8/AC

=> AC = 16/√3

Again tan 30 = AB/BC

=> 1/√3 = AB/8

=> AB = 8/√3

So height of tree = AB + AC

= 8/√3 + 16/√3

= 24/√3 m

Step-by-step explanation:

Answer 2

Let length of tree before windstorm is BD.

After windstorm the upper part of tree C falls from point C to point A on the ground

. Now, let CD = AC = h2 m AB = 10 m

Broken part makes an angle 60° from the ground.

So, ∠CAB = 60°

From right angled ∆ABC tan 60° = BC/AB ⇒ √3 = h1/10 ⇒ h1 = 10√3 and cos 60° = AB/AC ⇒ 1/2 = 10/h2 ⇒ h2 = 10 × 2 = 20 m

Hence, total length of tree BD = BC + CD = h1 + h1

= 10√3 + 20

= 10 × 1.732 + 20

= 17.32 + 20

= 37.32 m

Hence, height of the tree 37.32 m.