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Climate change and global warming are influencing storm behaviour, particularly in terms of intensity and rainfall. Strong winds and storms often cause uprooting and/or breaking of trees, which damage the vehicles standing underneath the trees.
On a particular day, during a high intensity storm, a tree broke such that its broken part formed an angle of $30^\circ$ with the ground. The distance between the base of the tree to the point where the top touches the ground is found to be 10 m.
Based on the above information, answer the following questions :
(i) Represent the given information with the help of a neat and well labelled diagram.
(ii) Find the height above the ground at which the tree is broken.
(iii) (a) Find the height of the tree before it broke. (Use $\sqrt{3} = 1.732$)
OR
(iii) (b) If another tree broke from the same height as in part (ii), but the broken part made a $60^\circ$ angle with the ground, find the total height of the tree.
On a particular day, during a high intensity storm, a tree broke such that its broken part formed an angle of $30^\circ$ with the ground. The distance between the base of the tree to the point where the top touches the ground is found to be 10 m.
Based on the above information, answer the following questions :
(i) Represent the given information with the help of a neat and well labelled diagram.
(ii) Find the height above the ground at which the tree is broken.
(iii) (a) Find the height of the tree before it broke. (Use $\sqrt{3} = 1.732$)
OR
(iii) (b) If another tree broke from the same height as in part (ii), but the broken part made a $60^\circ$ angle with the ground, find the total height of the tree.
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Solution: (i) Let B be the point at which tree is broken.
For Correct Figure (1 Mark)
(ii) In $\triangle BAC$
$\tan 30^\circ = \frac{x}{10}$
$x = \frac{10}{\sqrt{3}} \text{ m}$ or $\frac{10\sqrt{3}}{3} \text{ m}$ or $5.77 \text{ m}$ (1 Mark)
(iii) (a) In $\triangle BAC$
$\cos 30^\circ = \frac{10}{y}$
$y = \frac{20}{\sqrt{3}} \text{ m} = \frac{20\sqrt{3}}{3} \text{ m}$
$\therefore$ The height of tree before it broke = $x + y = \frac{10\sqrt{3}}{3} + \frac{20\sqrt{3}}{3} = 10\sqrt{3} = 17.32 \text{ m}$ (1 Mark)
OR
(b)
$\sin 60^\circ = \frac{x}{z}$
$z = \frac{20}{\sqrt{3}} \text{ m}$
$\therefore$ The height of tree before it broke = $(z + \frac{10}{\sqrt{3}}) \text{ m} = \frac{10}{3} (2 + \sqrt{3}) \text{ m}$ or $12.44 \text{ m}$ (1 Mark)
For Correct Figure (1 Mark)
(ii) In $\triangle BAC$
$\tan 30^\circ = \frac{x}{10}$
$x = \frac{10}{\sqrt{3}} \text{ m}$ or $\frac{10\sqrt{3}}{3} \text{ m}$ or $5.77 \text{ m}$ (1 Mark)
(iii) (a) In $\triangle BAC$
$\cos 30^\circ = \frac{10}{y}$
$y = \frac{20}{\sqrt{3}} \text{ m} = \frac{20\sqrt{3}}{3} \text{ m}$
$\therefore$ The height of tree before it broke = $x + y = \frac{10\sqrt{3}}{3} + \frac{20\sqrt{3}}{3} = 10\sqrt{3} = 17.32 \text{ m}$ (1 Mark)
OR
(b)
$\sin 60^\circ = \frac{x}{z}$
$z = \frac{20}{\sqrt{3}} \text{ m}$
$\therefore$ The height of tree before it broke = $(z + \frac{10}{\sqrt{3}}) \text{ m} = \frac{10}{3} (2 + \sqrt{3}) \text{ m}$ or $12.44 \text{ m}$ (1 Mark)