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प्रश्न
A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.
उत्तर
Let AC be the original height of the tree. Suppose BD be the broken part of the tree which is rested at D from the base of the tree.
Here, CD = 20 m and ∠BDC = 60º.
In right ∆BCD,
\[\tan60^\circ = \frac{BC}{CD}\]
\[ \Rightarrow \sqrt{3} = \frac{BC}{20}\]
\[ \Rightarrow BC = 20\sqrt{3} m . . . . . \left( 1 \right)\]
Also,
\[\cos60^\circ = \frac{CD}{BD}\]
\[ \Rightarrow \frac{1}{2} = \frac{20}{BD}\]
\[ \Rightarrow BD = 40 m . . . . . \left( 2 \right)\]
∴ Height of the tree = AB + BC = BD + BC =
\[\left( 40 + 20\sqrt{3} \right) m\] [Using (1) and (2)]
Thus, the height of the tree is \[\left( 40 + 20\sqrt{3} \right) m\]
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