Question

A tree breks due to Strom and broken part bends so that the top of the tree toches the ground by making an angle 45°with the ground the distance between the foot of the tree and the top of the tree on the ground is 8m find the height of the tree

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A tree breaks due to strom and broken part bends so that the top of the tree toches the ground by making an angle 45° with the ground and the distance between the foot of the tree and the top of the tree on the ground is 8m.

Let assume that, AB be the total height of tree.

Let assume that at the point C, the storm breaks the tree and broken part CB touches the ground at point D such that AD = 8 m

Now, CB = CD

Now, further given that ∠CDA = 45°.

Now, In right-angle triangle CDA

$\rm \: tan45 \degree \: = \dfrac{CA}{DA}$

$\rm \: 1 \: = \dfrac{CA}{8}$

$\rm\implies \:CA \: = \: 8 \: m \:$

Now, Again in right-angle triangle CDA

$\rm \: cos45 \degree \: = \: \dfrac{DA}{CD}$

$\rm \: \dfrac{1}{ \sqrt{2} } \: = \: \dfrac{8}{CD}$

$\rm\implies \:CD \: = \: 8 \sqrt{2} \: m$

So, Total height of tree is

$\rm \: =  \:AB$

$\rm \: =  \:AC + BC$

$\rm \: =  \:AC + CD$

$\rm \: = \: 8 + 8 \sqrt{2}$

$\rm \: =  \:8(1 + \sqrt{2})$

$\rm \: =  \:8(1 + 1.414)$

$\rm \: =  \:8 \times 2.414$

$\rm \: =  \:19.312 \: m$

$\rm\implies \:Total \: height \: of \: tree = AB \: = \: 19.312 \: m$

$\rule{190pt}{2pt}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$