Question

CLASS X
CHAPTER- Trigonometry


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.



➝ ANSWER WITH FULL EXPLANATION ​

Answer 1

Let the broken part of tree be AC

It is given that, distance between foot of the tree B and point C=8m.So, BC=8m

Also, broken parts of tree makes an angle 30 ∘

with ground.

So, ∠C=30 ∘

We need to find height of the tree

Height of the tree=Height of broken part+Height of remaining tree=AB+AC

Since, tree was vertical to ground. So, ∠ABC=90∘

In rightangled △ABC,

cosC= BC/ AC

⇒cos30 ∘ = 8/ AC

⇒

$\frac{ \sqrt{3} }{2} = \frac{8}{ac}$

= AC

$\frac{8 \times 2}{ \sqrt{3} } = \frac{16}{ \sqrt{3} }$

In rightangled △ABC,

sinC= AB/ AC

⇒sin30 ∘

= AB=

$\frac{1}{2} = \frac{16}{2 \sqrt{3} } = \frac{8}{ \sqrt{3} }$

So, height of the tree=AC+AB=

$\frac{16}{ \sqrt{3} } + \frac{8}{ \sqrt{3} } = \frac{24}{ \sqrt{3} }$

=

$\frac{24}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{24 \sqrt{3} }{3} = 8 \sqrt{3} m$

.

.

.

꧁Hope its help you꧂

Answer 2

$\LARGE{ \underline{\underline{\sf{Question:}}}}$

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.

$\LARGE{ \underline{\underline{\sf{Answer:}}}}$

$\sf{\:\:\:\:\:\:Refer \:the \:attachment \:for \:the \:figure}$

${ \underline{\underline{\sf{Understanding \:the \:problem:}}}}$

A tree is break due to storm . Tree is in straight line ,now its bend to the ground and form Angle of elevation.

Angle C = 30° . Where the tree is bend it form the distance btw the bend part and to the ground. CB = 8m

Here now need to find height AB = ?

${ \underline{\underline{\sf{Given}}}}$

$\sf{⇝\:AB \:be \:the \:actual \:height \:of \:the \:tree = ?}$

$\sf{⇝AC \:is \:the \:broken \:part \:touching \:the \:ground \:at \:D (AC = CD)}$

$\sf{⇝BD \:is \:bend \:part = 8m}$

${ \underline{\underline{\sf{Solution}}}}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:To \:find \:CB}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ΔCBD}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ \:tan 30° = \frac{Opposite}{Adjacent}}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ \frac{1}{\sqrt{3}} = \frac{CB}{8}}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ {\sqrt{3}}CB = 8}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ CB = \frac{8}{\sqrt{3}}m}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:To \:find \:CD}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ΔCBD}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ \:cos 30° = \frac{adjacent}{hypotenuse}}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ \frac{\sqrt{3}}{2} = \frac{8}{CD}}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ {\sqrt{3}}CD = 16}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ CD = \frac{16}{\sqrt{3}}m}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝AB = AC + CB}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝AB = \frac{16}{\sqrt{3}} + \frac{8}{\sqrt{3}}\:\:\:\:(AC = CD)}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ AB = \frac{24}{\sqrt{3}}}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝AB = \frac{24}{\sqrt{3}} + \frac{\sqrt{3}}{\sqrt{3}}}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ AB = \frac{24\sqrt{3}}{3}}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ AB = \frac{\cancel{24}^8 \sqrt{3}}{\cancel{3}}}$

$\sf{\:\:\:\:\:\:\:\:\:\:\:\:⇝ AB = 8\sqrt{3}m}$

$\large\sf\boxed{\:\:\:\:\:\:\:\:\:\:\:\:Height = AB = 8\sqrt{3}m}$