the upper part of a tree is broken by the wind and makes an angle of 30° with the ground. the distance from the foot of the point where the top touches the ground is 5m . the height of the tree is
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Answer:
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Let AB be the tree broken at a point C such that the broken part CB takes the position CO and strikes the ground at O.
It is given that OA = 25 m.
and <AOC = 30°.
Let AC = x and CB = y .
Then, CO = y .
\begin{gathered}In \triangle OAC ,we \:have\\ < /p > < p > tan 30\degree = \frac{AC}{OA}\\\implies \frac{1}{\sqrt{3}}=\frac{x}{25}\end{gathered}
In△OAC,wehave
</p><p>tan30°=
OA
AC
⟹
3
1
=
25
x
\implies x = \frac{25}{\sqrt{3}}\: ---(1)⟹x=
3
25
−−−(1)
\begin{gathered}And \\cos 30\degree = \frac{OA}{OC}\end{gathered}
And
cos30°=
OC
OA
\implies \frac{\sqrt{3}}{2}=\frac{25}{y}⟹
2
3
=
y
25
\implies y = \frac{50}{\sqrt{3}}\:---(2)⟹y=
3
50
−−−(2)
\begin{gathered}Height \: of \: the \: tree\\ = (x+y) \: m\\= \frac{25}{\sqrt{3}}+\frac{50}{\sqrt{3}}\\=\frac{25+50}{\sqrt{3}}\\=\frac{75}{\sqrt{3}}\\=\frac{75\sqrt{3}}{\sqrt{3}\times\sqrt{3}}\\=\frac{75\sqrt{3}}{3}\\=25\sqrt{3}\\=25\times 1.732\\= 43.3 \:m\end{gathered}
Heightofthetree
=(x+y)m
=
3
25
+
3
50
=
3
25+50
=
3
75
=
3
×
3
75
3
=
3
75
3
=25
3
=25×1.732
=43.3m
Therefore,
Height \: of \: the \: tree=43.3\:m
Height of the tree=43.3m