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Mathematics

A straight tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. The distance from the foot of the tree to the point where the top touches the ground is 10 metres. The height of the tree is :

  1. 10(3+1)10(\sqrt{3} + 1) m

  2. 10310\sqrt{3} m

  3. 10(3−1)10(\sqrt{3} - 1) m

  4. (103)\Big(\dfrac{10}{\sqrt{3}}\Big) m

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Answer

A straight tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with the ground. The distance from the foot of the tree to the point where the top touches the ground is 10 metres. The height of the tree is : Volume And Surface Area of solid RSA Mathematics Solutions ICSE Class 10.

Let the tree before breaking be OP.

The tree breaks at point C. The top part falls and touches the ground at point B.

Let BC = h1 and CP = h2

In triangle CPB,

⇒tan⁡(30∘)=CPBP⇒13=h210⇒h2=103 m.\Rightarrow \tan(30^\circ) = \dfrac{CP}{BP} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{h2}{10} \\[1em] \Rightarrow h2 = \dfrac{10}{\sqrt3} \text{ m.}

In triangle ABC,

⇒cos⁡(30∘)=BPBC⇒32=10h1⇒h1=203 m.\Rightarrow \cos(30^\circ) = \dfrac{BP}{BC} \\[1em] \Rightarrow \dfrac{\sqrt3}{2} = \dfrac{10}{h1} \\[1em] \Rightarrow h1 = \dfrac{20}{\sqrt3} \text{ m.}

The original height of the tree

=103+203=303=303×33=103.= \dfrac{10}{\sqrt3} + \dfrac{20}{\sqrt3} \\[1em] = \dfrac{30}{\sqrt3} \\[1em] = \dfrac{30}{\sqrt3} \times \dfrac{\sqrt3}{\sqrt3} \\[1em] = 10\sqrt3.

Hence, option 2 is the correct option.

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