The upper part of a tree broken over by the wind makes an angle of $30^{\circ}$ with ground, and the distance from the foot to the point where the top of the tree touches the ground is 15 m. the height of the tree is-

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Solution

R.E.F image
$c o s 30^{\circ} = \frac{\sqrt{3}}{2}$
$\frac{15}{B C} = \frac{\sqrt{3}}{2}$
$B C = 10 \sqrt{3}$
$T a n 30 = \frac{1}{\sqrt{3}}$ So, $A B + B C = 15 \sqrt{3}$
$\frac{A B}{B C} = \frac{1}{\sqrt{3}} \Rightarrow A B = 5 \sqrt{3}$

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The upper part of a tree broken over by the wind makes an angle of $30^{\circ}$ with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. Using sine rule, find the height of the tree.

Q. A tree

A C

is broken over by over by wind from

B . D

is the point where the top of the broken tree touches the ground and

B D

makes an angle of

45^{\circ}

with the ground, if the distance between he base of the tree and the point

D = 10

m. What is the height of the tree ?

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The upper part of a tree broken over by the wind makes an angle of 30∘ with ground, and the distance from the foot to the point where the top of the tree touches the ground is 15 m. the height of the tree is-

R.E.F image cos30∘=√3215BC=√32BC=10√3Tan30=1√3 So, AB+BC=15√3ABBC=1√3⇒AB=5√3

The upper part of a tree broken over by the wind makes an angle of $30^{\circ}$ with ground, and the distance from the foot to the point where the top of the tree touches the ground is 15 m. the height of the tree is-

R.E.F image
$c o s 30^{\circ} = \frac{\sqrt{3}}{2}$
$\frac{15}{B C} = \frac{\sqrt{3}}{2}$
$B C = 10 \sqrt{3}$
$T a n 30 = \frac{1}{\sqrt{3}}$ So, $A B + B C = 15 \sqrt{3}$
$\frac{A B}{B C} = \frac{1}{\sqrt{3}} \Rightarrow A B = 5 \sqrt{3}$