Question

A tree breaks down due to strim and the broken part bends so that the tip 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 8m find the height of tree

Answer 1

Question:

A tree breaks down due to strom 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 8m. Find the height of tree.

Answer:

■ Refer the image first.

●Let BC ne the unbroken part of the tree , AC be the broken part of tree which makes an angle of

30° with the ground.

●Such that ∠CAB =30° and ∠ABC= 90°.

● Height of the tree is BC+AC and AB= 8 m

●We need to Find the height of the tree.

In triangle ABC,

$\frac{p}{b} = tan \theta \\ \\ = > \frac{bc}{ab} = tan\theta \\ \\ = > \frac{bc}{8} = \frac{1}{ \sqrt{3} } \\ \\ = > \sqrt{3} \: bc = 8 \\ \\ = > bc = \frac{8}{ \sqrt{3} }$

Rationalising the denominator

$bc = \frac{8}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ \\ bc = \frac{8 \sqrt{3} }{3}$

Again in triangle ABC,

$= > \frac{p}{h} = sin\theta \\ \\ = > \frac{bc}{ac} = sin30 \degree \\ \\ = > \frac{ \frac{8 \sqrt{3} }{3} }{ac} = \frac{1}{2} \\ \\ = > ac = \frac{2 \times 8 \sqrt{3} }{3} \\ \\ = > ac = \frac{16 \sqrt{3} }{3}$

Total height of the tree = AC+BC

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

Hence the total height of the tree is √3