Ex.9.1 Q2 Some Applications of Trigonometry Solution - NCERT Maths Class 10

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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 \(30^{\circ}\) with it. The distance between the foot of the tree to the point where the top touches the ground is \(8\,\rm{m}\). Find the height of the tree.

Text Solution

  

What is Known?

(i) Broken part of the tree bends and touching the ground making an angle of \(30^{\circ}\) with the ground.

(ii) Distance between foot of the tree to the top of the tree is \(8\,\rm{m.}\)

What is Unknown?

Height of the tree

 

Reasoning:

(i) Height of the tree \( = AB + AC.\)

(ii) Trigonometric ratio which involves \(AB, BC\) and \(\angle C\) is \(\tan \theta\) , where \(AB\) can be measured.

(iii) Trigonometric ratio which involves \(AB, AC\) and \(\angle C\) is \(\sin \theta\) , where \(AC\) can be measured.

(iv) Distance between the foot of the tree to the point where the top touches the ground \(= BC = 8\,\rm{ m}\)

Steps:

In \(\Delta ABC\),

\[\begin{align}\tan C &=\frac{A B}{B C} \\ \tan 30^{\circ} &=\frac{A B}{8} \\ \frac{1}{\sqrt{3}} &=\frac{A B}{8} \\ A B &=\frac{8}{\sqrt{3}} \mathrm{m} \end{align}\]

\[\begin{align}\sin \,C{\rm{ }} &= {\rm{ }}\frac{{AB}}{{AC}}\\\sin 30^\circ  &= \left( {\frac{{\frac{8}{{\sqrt 3 }}}}{{AC}}} \right)\\\frac{1}{2}{\rm{ }} &= {\rm{ }}\frac{8}{{\sqrt 3 }} \times \frac{1}{{AC}}\\AC{\rm{ }} &= {\rm{ }}\frac{8}{{\sqrt 3 }} \times 2\\AC{\rm{ }} &= {\rm{ }}\frac{{16}}{{\sqrt 3 }}\,\end{align}\]

\[\begin{align}\text{Height of tree}\, &= \rm{ }AB + AC\\&= \frac{8}{{\sqrt 3 }}+ \frac{{16}}{{\sqrt 3 }}\\&= \frac{{24}}{{\sqrt 3 }}\\&= \frac{{24}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}\\&= \frac{{24\sqrt 3 }}{3}\\&= 8\sqrt 3 \end{align}\]

Height of tree \( =8 \sqrt{3} \rm{m}\)