Question

The shadow of a tower standing on a level ground is found to be 45 m longer when the
altitude (the angle of elevation) of sun changes from 60° to 30°. Find the height of the tower.
Or​

Answer 1

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

• Let AB be the height of tower.

and

• Let AC and AD be shadow of tower when angle of elevations are 60° and 30°.

So,

• CD = 45 m.

Let

• AB = 'h' meter

and

• AC = 'x' meter.

Now,

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

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

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

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

Now,

$\rm :\longmapsto\:In \: \triangle \: ABD$

$\rm :\longmapsto\:tan30\degree = \dfrac{AB}{AD}$

$\rm :\longmapsto\: \dfrac{1}{ \sqrt{3} } = \dfrac{h}{45 + x}$

$\rm :\longmapsto\: \dfrac{1}{ \sqrt{3} } = \dfrac{x \sqrt{3} }{45 + x} \: \: \: \: \: \: \: \: \: \: \: \red{ \because \{ h = x \sqrt{3} \}}$

$\rm :\longmapsto\:3x = x + 45$

$\rm :\longmapsto\:2x = 45$

$\rm :\implies\:x = 22.5 \: m$

On substituting the value of x in equation (1), we get

$\rm :\longmapsto\:h = 22.5 \times \sqrt{3}$

$\rm :\longmapsto\:h = 22.5 \times 1.732$

$\bf\implies \:h = 38.97 \: m$

$\overbrace{ \underline { \boxed { \bf \therefore \: The \: height \: of \: tower \: is \: 38.97 \: m}}}$

Additional Information :-

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