Question

A statue, 1.6 m 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

Answer:

answer is AB = 0.8 (√3+1)m

Answer 2

First Refer to the Diagram for the better understanding of question :-

Here Pedestal refers to the base of statue as given in the diagram.

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Assumption :-

• Let AB be the full height of the statue.

• CB be the height Pedestal.

Given :-

• Height above on the top of pedestal = AC = 1.6 m

• Angle ADB = 60° (Angle of Elevation)

• Angle CDB = 45° (Angle of Elevation of Pedestal)

To find :-

• Height of the Pedestal.

Solution :-

○ As we know that

${ \boxed{\sf{\ \tan( A )= \frac{ Perpendicular}{ Base}}}}$

○ In triangle CBD :-

${\sf{\implies{ \tan(A) = \cfrac{ Perpendicular}{ Base}}}}$

${\sf{\implies{ \tan(45 \degree) = \cfrac{CB}{ BD}}}}$

Here value of tan(A) is 1 so,

${\sf{\implies{1 = \cfrac{CB}{ BD}}}}$

${\sf{\implies{CB = BD}}}$

Let CB = BD be the equation 1 then,

${\sf{\implies{ CB = BD}}}...(eq \: 1)$

● In triangle ABD

${\sf{\implies{ \tan(A) = \cfrac{ Perpendicular}{ Base}}}}$

${\sf{\implies{ \tan(60 \degree) = \cfrac{ AC + CB}{ BD}}}}$

In equation 1 CB is equal to BD then,

${\sf{\implies{ \tan(60 \degree) = \cfrac{ AC + CB}{ CB}}}}$

Here value of tan(60°) is √3. So,

${\sf{\implies{ \sqrt{3} = \cfrac{ AC + CB}{CB}}}}$

${\sf{\implies{ \sqrt{3}CB = { AC + CB}}}}$

Here value of AC is 1.6 m because it is given.

${\sf{\implies{ \sqrt{3}CB = { 1.6 + CB}}}}$

${\sf{\implies{ \sqrt{3}CB -CB= { 1.6 }}}}$

Taking CB as common.

${\sf{\implies{ CB(\sqrt{3} -1)= { 1.6 }}}}$

${\sf{\implies{ CB= \cfrac { 1.6 }{(\sqrt{3} -1)} }}}$

Rationalizing the denominator:-

${\sf{\implies{ CB= \cfrac { 1.6 }{(\sqrt{3} -1)} \times \cfrac{( \sqrt{3} + 1)}{ ( \sqrt{3} + 1)} }}}$

Using formula [(a + b)(a - b) = a² - b²]

${\sf{\implies{ CB= \cfrac { 1.6 \times ( \sqrt{3} + 1)}{(\sqrt{3})^{2} -(1)^{2} } }}}$

${\sf{\implies{ CB= \cfrac { 1.6 \times ( \sqrt{3} + 1)}{3 - 1} }}}$

${\sf{\implies{ CB= \cfrac { 1.6 \times ( \sqrt{3} + 1)}{2} }}}$

${\sf{\implies{ CB= 0.8 \times ( \sqrt{3} + 1) }}}$

${ \boxed{ \red{\sf{\implies{ CB= 0.8 ( \sqrt{3} + 1) m }}}}}$

So, height of the Pedestal is 0.8(√3 + 1 ) m.

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More :- Further solving the value of CB.

${\sf{\rightarrow{CB = 0.8( \sqrt{3} + 1)}}}$

Considering the value of √3 as 1.73 then

${\sf{\rightarrow{CB = 0.8( 1.73+ 1)}}}$

${\sf{\rightarrow{CB = 0.8( 2.73)}}}$

${ \boxed{\sf{\rightarrow{CB =2.184 \: m }}}}$

By further solving we get the height of Pedestal as 2.184 m.

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More to know :-

$\begin{gathered}\footnotesize{\boxed{\begin{array}{c|c} \bf{\underline{ \: \:Ratio \: \: }} & \bf{\underline{ \: \: Formula \: \: }} \\ \\ \sf{ \sin(A) }&{ \sf{ \cfrac{ Perpendicular}{ Hypotenuse} }}\\ \\ \sf{ \cos(A) } & { \sf{\cfrac{Base }{ Hypotenuse} }} \\ \\ \sf{ \tan(A) } & \sf \cfrac{ Perpendicular }{Base } \\ \\ \sf{ \csc(A) } & \sf\cfrac{ Hypotenuse }{ Perpendicular } \\ \\ \sf{ \sec(A) } & \sf\cfrac{ Hypotenuse}{Base} \\ \\ \sf{ \cot(A) } & \sf \cfrac{Base}{Perpendicular} \: \\ \\ \end{array}}}\end{gathered}$