Question

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Answer 1

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Answer 2

Let the Height of the Pedestal be : AD

Let the Statue on the top of the Pedestal be : DB

Let the Point on the Ground be : C

Given : The Height of the Statue on the Pedestal = 1.6 m

$\implies \textsf{DB = 1.6 meter}$

Given : Angle of Elevation of the top of the Statue from point C is 60°

$\mathsf{\implies \angle BCA = 60^{\circ}}$

Given : Angle of Elevation of the top of the pedestal from point C is 60°

$\mathsf{\implies \angle DCA = 45^{\circ}}$

We know that :

$\bigstar\;\;\boxed{\mathsf{Tangent\;of\;a\;Triangle = \dfrac{Opposite\;Side\;of\;the\;Triangle}{Adjacent\;Side\;of\;the\;Traingle}}}$

From the Figure, Consider Triangle ADC :

In Triangle ADC :

★  AD is the Opposite side

★  AC is the Adjacent side

$\mathsf{\implies Tan45^{\circ} = \dfrac{AD}{AC}}$

$\mathsf{\implies AD = AC\;\;\;\;\;(since\;Tan45^{\circ} = 1)}$

Now, Consider Triangle ABC :

In Triangle ABC :

★  AB is the Opposite side

★  AC is the Adjacent side

$\mathsf{\implies Tan60^{\circ} = \dfrac{AB}{AC}}$

From the Figure, We can notice that : AB = AD + DB

$\mathsf{\implies \sqrt{3} = \dfrac{AD + DB}{AC} \;\;\;\;\;(since\;Tan60^{\circ} = \sqrt{3})}$

$\mathsf{\implies AC \times \sqrt{3} = AD + DB}$

But, We found that : AD = AC

$\mathsf{\implies AD \times \sqrt{3} = AD + DB}$

$\mathsf{\implies \sqrt{3}AD - AD = 1.6}$

$\mathsf{\implies AD(\sqrt{3} - 1) = 1.6}$

$\mathsf{\implies AD(1.732 - 1) = 1.6\;\;\;\;\;\;(since\;\sqrt{3} = 1.732)}$

$\mathsf{\implies AD(0.732) = 1.6}$

$\mathsf{\implies AD = \bigg(\dfrac{1.6}{0.732}\bigg)}$

$\mathsf{\implies AD = 2.186\;m}$

$\underline{\bf{Answer}} : \textsf{Height of the Pedestal is 2.186 meter}$