
A tree is broken at a certain height h its upper part $9\sqrt 2 $ long not completely separately meet the ground at angle of ${45^ \circ }$ , find the height of tree before it was broken and also find the distance from the root of the tree to the point where top of tree meets the ground?
Answer
574.8k+ views
Hint:From the information given in the question , Form a figure which will be a right angled triangle. Then by using the properties of the triangle and the trigonometric ratios solve the question.
Complete step-by-step answer:
Question says that the big tree is broken at some height which means after breaking it still has some height.
Let us assume that height to be $x$. Then the broken upper part is $9\sqrt 2 $ long and makes an angle of ${45^ \circ }$with the ground.
Therefore if we imagine the situation we may figure out that it forms a right angled triangle with perpendicular the broken tree base as the distance between the roots of the tree and top of the broken tree and hypotenuse as the broken part.
By using trigonometric ratios:
$
\tan \,{\theta } = \dfrac{{perpendicular}}{{base}} \\
\tan \,{45^{\circ} } = \dfrac{{perpendicular}}{{base}} \\
$
We know from trigonometric standard angles table $\tan 45^{\circ} = 1$
So substituting all the values we get,
$
\Rightarrow 1 = \dfrac{x}{{base}} \\
\Rightarrow base = x \\ $
This means that the height of the broken tree and the distance between the roots and the broken top is the same.
Since this is right angled triangle
Therefore by using Pythagoras theorem
\[
\Rightarrow {(height)^2} + {(base)^2} = {(hypotenuse)^2} \\
\Rightarrow {x^2} + {x^2} = {(9\sqrt 2 )^2} \\
\Rightarrow 2{x^2} = 2 \times 81 \\
\Rightarrow {x^2} = 81 \\
\Rightarrow x = \pm \sqrt {81} = \pm 9 \\
\]
Since, $x$ represents the height of the tree.
So $x = - 9$ is not possible
$ \Rightarrow x = 9$
Perpendicular = base = $9$m
Total height of tree $ = $ height of broken tree $ + $height of broken top
$ = 9 + 9\sqrt 2 $m
Distance from roots to broken top = base
$ = 9m$
Note:The most important part in solving these kind of questions is imagining the given situation and drawing the diagram according to information given in the question.Further Students should remember trigonometric ratios and Pythagoras theorem for solving these types of questions.
Complete step-by-step answer:
Question says that the big tree is broken at some height which means after breaking it still has some height.
Let us assume that height to be $x$. Then the broken upper part is $9\sqrt 2 $ long and makes an angle of ${45^ \circ }$with the ground.
Therefore if we imagine the situation we may figure out that it forms a right angled triangle with perpendicular the broken tree base as the distance between the roots of the tree and top of the broken tree and hypotenuse as the broken part.
By using trigonometric ratios:
$
\tan \,{\theta } = \dfrac{{perpendicular}}{{base}} \\
\tan \,{45^{\circ} } = \dfrac{{perpendicular}}{{base}} \\
$
We know from trigonometric standard angles table $\tan 45^{\circ} = 1$
So substituting all the values we get,
$
\Rightarrow 1 = \dfrac{x}{{base}} \\
\Rightarrow base = x \\ $
This means that the height of the broken tree and the distance between the roots and the broken top is the same.
Since this is right angled triangle
Therefore by using Pythagoras theorem
\[
\Rightarrow {(height)^2} + {(base)^2} = {(hypotenuse)^2} \\
\Rightarrow {x^2} + {x^2} = {(9\sqrt 2 )^2} \\
\Rightarrow 2{x^2} = 2 \times 81 \\
\Rightarrow {x^2} = 81 \\
\Rightarrow x = \pm \sqrt {81} = \pm 9 \\
\]
Since, $x$ represents the height of the tree.
So $x = - 9$ is not possible
$ \Rightarrow x = 9$
Perpendicular = base = $9$m
Total height of tree $ = $ height of broken tree $ + $height of broken top
$ = 9 + 9\sqrt 2 $m
Distance from roots to broken top = base
$ = 9m$
Note:The most important part in solving these kind of questions is imagining the given situation and drawing the diagram according to information given in the question.Further Students should remember trigonometric ratios and Pythagoras theorem for solving these types of questions.
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