Question
A tree breaks and falls to the ground such that its upper part is still partially attached to its stem. At what height did it break, if the original height of the tree was 24 cm and it makes an angle of 30° with the ground?
Answer: Option B
Let the tree break at height h cm from ground at point M.
The broken part makes angle of 30°
∴ Broken Part MP = MN = 24 - h
$$\eqalign{
& {\text{in}}\,\Delta MCQ, \cr
& \sin {30^ \circ } = \frac{1}{2} = \frac{{MQ}}{{MN}} = \frac{{\text{h}}}{{24 - {\text{h}}}} \cr
& \therefore 24 - {\text{h}} = 2{\text{h}} \cr} $$
∴ h = 8 cm = Tree breaks at this height
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Let the tree break at height h cm from ground at point M.
The broken part makes angle of 30°
∴ Broken Part MP = MN = 24 - h
$$\eqalign{
& {\text{in}}\,\Delta MCQ, \cr
& \sin {30^ \circ } = \frac{1}{2} = \frac{{MQ}}{{MN}} = \frac{{\text{h}}}{{24 - {\text{h}}}} \cr
& \therefore 24 - {\text{h}} = 2{\text{h}} \cr} $$
∴ h = 8 cm = Tree breaks at this height
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