Question

I am giving 50 points please give me proper explaination ​

Answer 1

${ \underline{ \large{ \pmb{ \sf{Solution:}}}}}$

${ \sf{Let \: the \: initial \: height \: of \: the \: tree \: be \: AC}}$

${ \sf{When \: the \: storm \: come, \: the \: tree \: broke \: from \: point \: B}}$

${ \sf{The \: broken \: part \: of \: the \: tree \: BC \: touches \: the \: ground \: at \: D}}$

${ \sf{it \: makes \: an \: angle \: 30° \: on \: the \: ground}}$

Also, AD = 6m

In, Right Angle triangle BAD

$: { \implies{ \sf{tan30 = \frac{AB}{AD} }}} \\ \\ : { \implies{ \sf{ \frac{1}{ \sqrt{3} } = \frac{AB}{6} }}} \\ \\ : { \implies{ \sf{AB = \frac{6}{ \sqrt{3} } m}}}$

In triangle, BAD,

$: { \implies{ \sf{cos30 = \frac{AD}{BD} }}} \\ \\ : { \implies{ \sf{ \frac{ \sqrt{3} }{2} = \frac{6}{BD} }}} \\ \\ : { \implies{ \sf{BD = \frac{12}{ \sqrt{3} }m }}}$

But, BD = BC

• AC = AB + BC

$: { \implies{ \sf{ \frac{6}{ \sqrt{3} } + \frac{12}{ \sqrt{3} } }}} \\ \\ : { \implies{ \sf{ \frac{18}{ \sqrt{3} } }}} \\ \\ : { \implies{ \sf{ \frac{18}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } }}} \\ \\ : { \implies{ \sf{ \frac{18 \sqrt{3} }{3} }}} \\ \\ : { \implies{ \sf{6 \sqrt{3} }}}$

${ \therefore{ \large{ \pmb{ \sf{Height \: of \: tree \: before \: falling = 6 \sqrt{3} m}}}}}$