Upper part of a vertical tree which is broken over by the winds just touches the ground and makes an angle of $30^{0}$ with the ground. If the length of the broken part is 20 metros, then the remaining part of the tree is of length

20 metres

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10 \sqrt{3}

metres

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10 metres

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10 \sqrt{2}

metres

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Solution

The correct option is A 10 metres
$T h e t r e e b r e a k s a t P . ∴ P O = 20 m, ∠ P O B = 30^{\circ} ∴ \frac{P B}{P O} = s i n 30^{\circ} o r \frac{P B}{20} = \frac{1}{2} o r P B = \frac{1}{2} \times 20 = 10 m (A n s)$ .

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Similar questions

Q. Upper part of a vertical tree which is broken over by the winds just touches the ground and makes an angle of

30^{o}

with the ground. If the length of the broken part is

20

metres, then the remaining part of the tree is of length.

Q. Upper part of a vertical tree which is broken over by the winds just touches the ground and makes as angle of

30^{\circ}

with the ground if the length of the broken part is 20 metres then remaining part of the tree is of length

A tree is broken by wind and its upper part touches the ground at a point $10$ metres from the foot of the tree and makes an angle of $45^{\circ}$ with the ground. The entire length of the tree is

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Upper part of a vertical tree which is broken over by the winds just touches the ground and makes an angle of 300 with the ground. If the length of the broken part is 20 metros, then the remaining part of the tree is of length

The correct option is A 10 metresThetreebreaksatP.∴PO=20m,∠POB=30∘∴PBPO=sin30∘orPB20=12orPB=12×20=10m(Ans).

Upper part of a vertical tree which is broken over by the winds just touches the ground and makes an angle of $30^{0}$ with the ground. If the length of the broken part is 20 metros, then the remaining part of the tree is of length

The correct option is A 10 metres
$T h e t r e e b r e a k s a t P . ∴ P O = 20 m, ∠ P O B = 30^{\circ} ∴ \frac{P B}{P O} = s i n 30^{\circ} o r \frac{P B}{20} = \frac{1}{2} o r P B = \frac{1}{2} \times 20 = 10 m (A n s)$ .