Question

A statue 1.6m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°.Find the height of the pedestal​

Answer 1

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

Given that,

A statue 1.6m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°.

Let assume that AB represents the pedestal and Let assume that BC be the 1.6 m tall statue.

Let assume that D be any point on the ground such that the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°.

So, it means ∠ADB = 45° and ∠ADC = 60°.

Let further, Assume that height of pedestal be 'h' m.

So, AB = h m.

Further, assume that AD = x m.

Now, In right angle triangle ADB

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

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

$\rm \implies\:\boxed{ \tt{ \: \: x \: = \: h \: \: }} - - - - (1)$

Now, In right angle triangle ACD

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

$\rm :\longmapsto\: \sqrt{3} = \dfrac{AB + BC}{AD}$

$\rm :\longmapsto\: \sqrt{3} = \dfrac{h + 1.6}{x}$

$\rm :\longmapsto\: \sqrt{3} = \dfrac{h + 1.6}{h}$

$\rm :\longmapsto\: \sqrt{3}h = 1.6 + h$

$\rm :\longmapsto\: \sqrt{3}h - h = 1.6$

$\rm :\longmapsto\: (\sqrt{3} - 1)h = 1.6$

$\rm :\longmapsto\:h = \dfrac{1.6}{ \sqrt{3} - 1}$

On rationalizing the denominator, we get

$\rm :\longmapsto\:h = \dfrac{1.6}{ \sqrt{3} - 1} \times \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} + 1}$

$\rm :\longmapsto\:h = \dfrac{1.6( \sqrt{3} + 1) }{ {( \sqrt{3} )}^{2} - {1}^{2} }$

$\rm :\longmapsto\:h = \dfrac{1.6( \sqrt{3} + 1) }{3 - 1}$

$\rm :\longmapsto\:h = \dfrac{1.6( \sqrt{3} + 1) }{2}$

$\rm :\longmapsto\:h = 0.8( \sqrt{3} + 1) \: m$

or

$\rm :\longmapsto\:h = 0.8(1.732 + 1) \: m$

$\rm :\longmapsto\:h = 0.8(2.732) \:$

$\bf\implies \:h \: = \: 2.1856 \: m$

Hence,

$\rm \implies\:\boxed{ \tt{ \: h \: = \: 0.8( \sqrt{3} + 1) \: m \: or \: 2.1856 \: m \: }}$

More to know :-

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