Question

the upper part of a tree breaks due to windstorm and it makes an angle of 60 with the ground .the distance from the bottom of the tree to the point. where the top touches the ground is 10m .find the original height of the tree
$\sqrt{3 = 1.732}$
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Answer 1

let AB be the tree of height[ h] m now let the tree be broken from point C which touches the ground at point D which is 10 m apart from the base of the tree.the abgle of elevation of tree with the ground is 60.

let the height of broken tree be x m. We were to find the total height of the tree.

in triangle CBD

CB/BD=tan60

x/10 = root3

x = 10*root 3

now again in triangle CBD

CB/CD = sin 60

x/h = root 3/2

10/root3/h = root 3/2

20 = h

height of the tree = x+

=10*root 3+20

=1.732*10+20

= 17.32 m+20

height of the tree = 37.32 m   ANS

Answer 2

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 60 = BC/AC

=> 1/2 = 10/AC

=> AC = 20

Again tan 60 = AB/BC

 =>√3 = AB/10

=> AB = 10/√3

So height of tree = AB + AC

 = 10/√3 +20

= 37.32m Ans