[tex]\huge\fbox \red{h}\fbox \green{e}\fbox \purple{l}\fbox \orange{l}\fbox \red{o}[/tex] A tree breaks due to storm and the broken part bends so 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 8 m. Find the height of the tree.➳ REQUIRED GOOD ANSWER. Created by itsAyan. Math

tex] A tree breaks due to storm and the broken part bends so that the top of the...

$\large\underbrace{\underline{\sf{Understanding\; the\; Concept}}}}$

Here in this question concept of trigonometry is applied. We have given that a tree breaks due to storm and the broken part made 30° angle with the ground and the distance of broken part to distance of foot of tree is given to be 8 m and we have to find height of tree which could be find easily by applying trigonometry ratios.

So let's start!

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Diagram:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(2,2)(3,3)\thicklines\qbezier(3,0)(2,0)(0,0)\qbezier(3,0)(3,1)(3,3){\put(2.76,0.13){\framebox}}\put(1,-0.4){\sf8 metre}\put(0.4,0.1){30}\put(3.2,1.4){\sf x metre}\end{picture}$

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As it can be observed from the diagram that we have to use ratios related to height and base of triangle.

We have:

$\underline{\boxed{\mathfrak{ Tan\; \theta=\dfrac{Height}{Base}}}}\star$

Here angle will be 30° according to the question.

So:-

$\sf:\implies Tan\: 30^{\circ}=\dfrac{Height}{Base}$

$\sf:\implies \dfrac{1}{\sqrt3}=\dfrac{Height}{8\:metre}$

$\sf:\implies \dfrac{8\:metre}{\sqrt3}=Height$

Now rationalising it:

$\sf:\implies \dfrac{8\:metre}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=Height$

$\sf:\implies \dfrac{8\sqrt3}{3}=Height$

$\underline{\boxed{\mathfrak {2.66 \sqrt3 \:metre=Height}}}$

So the required height of the tree is 2.66√3 m.

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Trigonometric ratios:

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} $ \sin $ & 0 & $\dfrac{1}{2 }$ & $\dfrac{1}{ \sqrt{2} }$ & $\dfrac{ \sqrt{3}}{2}$ & 1 \\ \cline{1-6} $ \cos $ & 1 & $ \dfrac{ \sqrt{ 3 }}{2} } $ & $ \dfrac{1}{ \sqrt{2} } $ & $ \dfrac{ 1 }{ 2 } $ & 0 \\ \cline{1-6} $ \tan $ & 0 & $ \dfrac{1}{ \sqrt{3} } $ & 1 & $ \sqrt{3} $ & $ \infty $ \\ \cline{1-6} \cot & $ \infty $ &$ \sqrt{3} $ & 1 & $ \dfrac{1}{ \sqrt{3} } $ &0 \\ \cline{1 - 6} \sec & 1 & $ \dfrac{2}{ \sqrt{3}} $ & $ \sqrt{2} $ & 2 & $ \infty $ \\ \cline{1-6} \csc & $ \infty $ & 2 & $ \sqrt{2 } $ & $ \dfrac{ 2 }{ \sqrt{ 3 } } $ & 1 \\ \cline{1 - 6}\end{tabular}}$

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Kindly see this answer on web.

tejasvinisinhaps23

Let the Height of the Tree =AB+AD

and given that BD=8 m

Now, when it breaks a part of it will remain perpendicular to the ground (AB) and remaining part (AD) will make an angle of 30

Now, in △ABD

cos30

⇒BD=

⇒AD=

2×8

also, in the same Triangle

tan30

⇒AB=

∴ Height of tree =AB+AD=(

)m=

m=8

please mark me brainliest and follow

me darling

Questions

Answers & Comments

So let's start!

Diagram:

We have:

So:-

Now rationalising it:

So the required height of the tree is 2.66√3 m.

Trigonometric ratios:

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