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Hint: Use trigonometric ratios to find the unknown sides in the triangle and evaluate the equations to find the height of the tree.

First, let us draw a diagram to represent what’s given in the question.

AC is the broken base of the tree and BC is the tree which has bent and touched the ground making an angle of $30^\circ $ with the ground.

(AC+BC) is the total height of the tree which is asked.

AB is the distance between the foot of the tree and the point where the tree touches the ground.

AB=10 meters

Let AC=x meters and let BC=y meters

$\angle ABC = 30^\circ ,\angle CAB = 90^\circ $

Let us find AC and BC using trigonometric ratios.

To find AC, we will find $\tan 30^\circ $. The value of $\tan \theta = \dfrac{{opposite}}{{adjacent}}$

$

\tan 30^\circ = \dfrac{{AC}}{{AB}} \\

\dfrac{1}{{\sqrt 3 }} = \dfrac{x}{{10}} \\

x = \dfrac{{10}}{{\sqrt 3 }} \\

$

Hence, $AC = \dfrac{{10}}{{\sqrt 3 }}$meters

To find BC, we will find $\cos 30^\circ $. The value of $\cos \theta = \dfrac{{adjacent}}{{hypotenuse}}$

$

\cos 30^\circ = \dfrac{{AB}}{{BC}} \\

\dfrac{{\sqrt 3 }}{2} = \dfrac{{10}}{y} \\

y = \dfrac{{20}}{{\sqrt 3 }} \\

$

Hence,$BC = \dfrac{{20}}{{\sqrt 3 }}$meters

Height of the tree= (AC+BC)

\[

= \dfrac{{10}}{{\sqrt 3 }} + \dfrac{{20}}{{\sqrt 3 }} \\

= \dfrac{{30}}{{\sqrt 3 }} \\

= \dfrac{{3 \times 10}}{{\sqrt 3 }} \\

= 10\sqrt 3 \\

\]

Hence, the total height of the tree is $10\sqrt 3 $meters.

Note: Students tend to make the mistake of not adding AC and BC. Instead they give the answer as only the value of AC. The height of the tree means, before the tree was broken.

First, let us draw a diagram to represent what’s given in the question.

AC is the broken base of the tree and BC is the tree which has bent and touched the ground making an angle of $30^\circ $ with the ground.

(AC+BC) is the total height of the tree which is asked.

AB is the distance between the foot of the tree and the point where the tree touches the ground.

AB=10 meters

Let AC=x meters and let BC=y meters

$\angle ABC = 30^\circ ,\angle CAB = 90^\circ $

Let us find AC and BC using trigonometric ratios.

To find AC, we will find $\tan 30^\circ $. The value of $\tan \theta = \dfrac{{opposite}}{{adjacent}}$

$

\tan 30^\circ = \dfrac{{AC}}{{AB}} \\

\dfrac{1}{{\sqrt 3 }} = \dfrac{x}{{10}} \\

x = \dfrac{{10}}{{\sqrt 3 }} \\

$

Hence, $AC = \dfrac{{10}}{{\sqrt 3 }}$meters

To find BC, we will find $\cos 30^\circ $. The value of $\cos \theta = \dfrac{{adjacent}}{{hypotenuse}}$

$

\cos 30^\circ = \dfrac{{AB}}{{BC}} \\

\dfrac{{\sqrt 3 }}{2} = \dfrac{{10}}{y} \\

y = \dfrac{{20}}{{\sqrt 3 }} \\

$

Hence,$BC = \dfrac{{20}}{{\sqrt 3 }}$meters

Height of the tree= (AC+BC)

\[

= \dfrac{{10}}{{\sqrt 3 }} + \dfrac{{20}}{{\sqrt 3 }} \\

= \dfrac{{30}}{{\sqrt 3 }} \\

= \dfrac{{3 \times 10}}{{\sqrt 3 }} \\

= 10\sqrt 3 \\

\]

Hence, the total height of the tree is $10\sqrt 3 $meters.

Note: Students tend to make the mistake of not adding AC and BC. Instead they give the answer as only the value of AC. The height of the tree means, before the tree was broken.