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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 $10 \text{ m}$.
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 $10 \text{ m}$.
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 $\Delta BAC$,
$\tan 30^{\circ} = \frac{x}{10}$
$x = \frac{10}{\sqrt{3}} \text{ m}$ or $\frac{10\sqrt{3}}{3} \text{ m}$ or $5.77 \text{ m}$ (1 Mark)
$:::$ The height above the ground at which the tree is broken is $\frac{10}{\sqrt{3}} \text{ m}$ or $\frac{10\sqrt{3}}{3} \text{ m}$ or $5.77 \text{ m}$
(iii) (a) In $\Delta BAC$,
$\cos 30^{\circ} = \frac{10}{y}$
$y = \frac{10}{\cos 30^{\circ}} = \frac{10}{\sqrt{3}/2} = \frac{20}{\sqrt{3}} \text{ m}$ or $\frac{20\sqrt{3}}{3} \text{ m}$ (1 Mark)
$:::$ The height of tree before it broke $= x + y = \frac{10}{\sqrt{3}} + \frac{20}{\sqrt{3}} = \frac{30}{\sqrt{3}} = 10\sqrt{3} = 17.32 \text{ m}$ (1 Mark)
OR
(iii) (b)
$\sin 60^{\circ} = \frac{10}{z}$
$z = \frac{10}{\sin 60^{\circ}} = \frac{10}{\sqrt{3}/2} = \frac{20}{\sqrt{3}} \text{ m}$ (1 Mark)
The height of tree before it broke $= (z + \frac{10\sqrt{3}}{3}) \text{ m} = (\frac{20}{\sqrt{3}} + \frac{10}{\sqrt{3}}) \text{ m} = \frac{30}{\sqrt{3}} = 10\sqrt{3} \text{ m}$ or $17.32 \text{ m}$ (1 Mark)
(ii) In $\Delta BAC$,
$\tan 30^{\circ} = \frac{x}{10}$
$x = \frac{10}{\sqrt{3}} \text{ m}$ or $\frac{10\sqrt{3}}{3} \text{ m}$ or $5.77 \text{ m}$ (1 Mark)
$:::$ The height above the ground at which the tree is broken is $\frac{10}{\sqrt{3}} \text{ m}$ or $\frac{10\sqrt{3}}{3} \text{ m}$ or $5.77 \text{ m}$
(iii) (a) In $\Delta BAC$,
$\cos 30^{\circ} = \frac{10}{y}$
$y = \frac{10}{\cos 30^{\circ}} = \frac{10}{\sqrt{3}/2} = \frac{20}{\sqrt{3}} \text{ m}$ or $\frac{20\sqrt{3}}{3} \text{ m}$ (1 Mark)
$:::$ The height of tree before it broke $= x + y = \frac{10}{\sqrt{3}} + \frac{20}{\sqrt{3}} = \frac{30}{\sqrt{3}} = 10\sqrt{3} = 17.32 \text{ m}$ (1 Mark)
OR
(iii) (b)
$\sin 60^{\circ} = \frac{10}{z}$
$z = \frac{10}{\sin 60^{\circ}} = \frac{10}{\sqrt{3}/2} = \frac{20}{\sqrt{3}} \text{ m}$ (1 Mark)
The height of tree before it broke $= (z + \frac{10\sqrt{3}}{3}) \text{ m} = (\frac{20}{\sqrt{3}} + \frac{10}{\sqrt{3}}) \text{ m} = \frac{30}{\sqrt{3}} = 10\sqrt{3} \text{ m}$ or $17.32 \text{ m}$ (1 Mark)