Question

From the top of a building 60m high, the angles of elevation and depression are found to be 60° and 30°, when the tip and base of a hill are observed respectively.Find the height of the hill​

Answer 1

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

Given that,

From the top of a building 60m high, the angles of elevation and depression are found to be 60° and 30°, when the tip and base of a hill are observed respectively.

Let assume that AB be the building whose height is 60 m and Let assume that CD be the hill and let height of the hill CD be h m.

Now, From A, draw AE perpendicular on CD intersecting CD at E.

Now, AB = DE = 60 m

So, CE = CD - DE = h - 60 m

Let assume that AE = x m

Now, In right angle triangle AED

$\rm \: tan30\degree = \dfrac{ED}{AE}$

$\rm \: \dfrac{1}{ \sqrt{3} } = \dfrac{60}{x}$

$\rm\implies \:x = 60 \sqrt{3} \: m$

Now, In right angle triangle AEC

$\rm \: tan60\degree = \dfrac{EC}{AE}$

$\rm \: \sqrt{3} = \dfrac{h - 60}{60 \sqrt{3} }$

$\rm \: h - 60 = 180$

$\rm \: h = 60 + 180$

$\rm\implies \:h \: = \: 240 \: m$

So, height of the hill is 240 m

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

Additional Information :-

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