Math, asked by dhruvbansal1862, 1 year ago

The angle of depression of the top and the bottom of an 8m tall building from the top of multi-storeyed building 30' and 45' respectively . Find the height of the multi-storeyed building and the distance between two building?

Answers

Answered by Anonymous
19

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Answered by mathdude500
0

Answer:

 \:\boxed{\begin{aligned}&  \:\sf \:Height\:of\:multi-storeyed\:building = 4(3 +  \sqrt{3}) \: m  \: \\ \\&  \:\sf \: Distance\:between\:two\:buildings=4(3 +  \sqrt{3}) \: m  \end{aligned}} \qquad \: \\  \\

Step-by-step explanation:

Let assume that AB represents the building and CD represents the multi-storeyed building such that AB = 8 m.

Now, from top C of the multi-storeyed building, the angle of depression of top and bottom of building AB be 30° and 45° respectively.

Let assume that CD = h m and BD = x m.

From A, draw AE perpendicular to CD intersecting CD at E.

Now, AE = BD = x m

So, CE = CD - DE = CD - AB = (h - 8) m

Now, In right-angle triangle CDB

\sf \: tan45 \degree \:  =  \: \dfrac{CD}{DB}  \\  \\

\sf \: 1 \:  =  \: \dfrac{h}{x}  \\  \\

\sf\implies \sf \: h = x \\  \\

Now, In right-angle triangle CEA

\sf \: tan30 \degree \:  =  \: \dfrac{CE}{EA}  \\  \\

\sf \:   \dfrac{1}{\sqrt{3}}  \:  =  \: \dfrac{h - 8}{x}  \\  \\

\sf \:   \dfrac{1}{ \sqrt{3} } \:  =  \: \dfrac{h - 8}{h}  \\  \\

\sf \:  \sqrt{3}(h - 8)  \:  =  \: h  \\  \\

\sf \:  \sqrt{3}h - 8 \sqrt{3}   \:  =  \: h  \\  \\

\sf \:  \sqrt{3}h - h = 8 \sqrt{3}  \\  \\

\sf \:  (\sqrt{3} -1) h = 8 \sqrt{3}  \\  \\

\sf \: h = \dfrac{8 \sqrt{3} }{ \sqrt{3} - 1 }  \\  \\

On rationalizing the denominator, we get

\sf \: h = \dfrac{8 \sqrt{3} }{ \sqrt{3} - 1 }   \times  \dfrac{ \sqrt{3}  + 1}{ \sqrt{3} + 1 }  \\  \\

\sf \: h = \dfrac{8 \sqrt{3} ( \sqrt{3}  + 1)}{( \sqrt{3})^{2}  -  {(1)}^{2} }  \\  \\

\sf \: h = \dfrac{8 \sqrt{3} ( \sqrt{3}  + 1)}{3 - 1 }  \\  \\

\sf \: h = \dfrac{8 \sqrt{3} ( \sqrt{3}  + 1)}{2}  \\  \\

\sf \: h = 4 \sqrt{3} ( \sqrt{3}  + 1)  \\  \\

\sf\implies \sf \: h = 4 (3 +  \sqrt{3}) \: m  \\  \\

So,

\sf\implies \sf \: x = 4 (3 +  \sqrt{3}) \: m  \\  \\

Hence,

 \:\boxed{\begin{aligned}&  \:\sf \:Height\:of\:multi-storeyed\:building = 4(3 +  \sqrt{3}) \: m  \: \\ \\&  \:\sf \: Distance\:between\:two\:buildings=4(3 +  \sqrt{3}) \: m  \end{aligned}} \qquad \: \\  \\

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