Question

A girl observes the angle of elevation of a mountain top to be 60° after walking directly away from it on level ground through 100m , the angle of elevation is 45°. Find the height of the mountain and the distance between the mountain and first position of the girl​

Answer 1

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

• Let the height of mountain AB be 'h' meter.

• Let the girl is at point, observe the angle of elevation of the top of mountain B as 60°.

• Let the distance between girl and foot of mountain in this position, AC = 'x' meter.

Now, the girl walk away 100 m from the point C.

• Let new position is at point D such that CD = 100 m.

and

• angle of elevation of the top of the mountain is 45°.

Now,

$\rm :\longmapsto\:In \: \triangle \: ABC$

$\rm :\longmapsto\:tan60 \degree \: = \dfrac{AB}{AC}$

$\rm :\longmapsto\: \sqrt{3} = \dfrac{h}{x}$

$\bf\implies \:h \: = \: \sqrt{ 3} x \: - - (1)$

Now,

$\rm :\longmapsto\:In \: \triangle \: BAD$

$\rm :\longmapsto\:tan45 \degree \: = \: \dfrac{BA}{AD}$

$\rm :\longmapsto\:1 \: = \: \dfrac{h}{100 + x}$

$\rm :\longmapsto\:100 + x = \sqrt{3} x \: \: \: \: \: \: \: \: \{using \: (1) \}$

$\rm :\longmapsto\: \sqrt{ 3} x - x = 100$

$\rm :\longmapsto\:x( \sqrt{3} - 1) = 100$

$\rm :\longmapsto\:x = \dfrac{100}{ \sqrt{3} - 1 } \times \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} + 1 }$

$\rm :\longmapsto\:x = \dfrac{100( \sqrt{3} + 1)}{3 - 1}$

$\rm :\longmapsto\:x = \dfrac{100( \sqrt{3} + 1) }{2}$

$\bf :\implies\:x = 50( \sqrt{3} + 1) \: meter- - - (2)$

On substituting value of x in equation (2), we get

$\rm :\longmapsto\:h \: = \: \sqrt{3} \times 50( \sqrt{3} + 1)$

$\bf\implies \:h \: = \: 50(3 + \sqrt{3} ) \: meter.$

Explore more :-

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