Math, asked by arts2566, 1 year ago

16. The angle of elevation of the top of a vertical tower from a point on the ground is 60o . From another point 10 m vertically above the first, its angle of elevation is 45o . Find the height of the tower.

Answers

Answered by riteshsingh77
1
solution in given in photo
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Answered by Anonymous
1

Answer:

Let the height be EC

Let the the angle EAD be 45° and the angle EBC be 60°

Let EB be x m and in reactangle ABCD we know that opposite sides are equal: AB = CD = 10 m and BC = AD

Now,

\sf In \:  \Delta  \: AED, \\

:\implies \sf tan \: 45^{\circ} = \dfrac{ED}{AD} \\  \\

:\implies \sf tan \: 45^{\circ} = \dfrac{ED}{BC} \\  \\

:\implies \sf tan \: 45^{\circ} = \dfrac{x}{BC} \\  \\

:\implies \sf 1= \dfrac{x}{BC} \\  \\

:\implies \sf x = BC \:\:  \:  \: \:\:\:\Bigg\lgroup \bf Equation\:(i)\Bigg\rgroup \\  \\  \\

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\sf In \:  \Delta  \: EBC, \\

:\implies \sf tan \: 60^{\circ} = \dfrac{EC}{BC} \\  \\

:\implies \sf  \sqrt{3} = \dfrac{x + 10}{BC} \\  \\

:\implies \sf  BC (\sqrt{3} ) = x + 10 \:\:  \:  \: \:\:\:\Bigg\lgroup \bf taking\:BC \: towards \:LHS \Bigg\rgroup \\  \\

:\implies \sf  x\sqrt{3} - x = 10 \\  \\

:\implies \sf  x \:  \left(\sqrt{3} - 1 \right) = 10 \\  \\

:\implies \sf x =  \dfrac{10}{( \sqrt{3} - 1) }  \:\:  \:  \: \:\:\:\Bigg\lgroup \bf Equation\:(ii)\Bigg\rgroup \\  \\  \\

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★ Height of the tower will be :

\dashrightarrow\:\: \sf Height= x+ 10 \\  \\

\dashrightarrow\:\: \sf  Height=   \frac{10}{ \sqrt{3 }  - 1} + 10 \\  \\

\dashrightarrow\:\: \sf  Height=  \dfrac{10 \sqrt{3}}{ \sqrt{3} - 1 } \\  \\

\dashrightarrow\:\: \sf  Height=15 + 5 \sqrt{3}  \\  \\

\dashrightarrow\:\: \sf  Height=15 + 5  \times 1.73 \:  \:  \:  \Bigg\lgroup \bf Putting \:  \sqrt{3} = 1.73 \Bigg\rgroup\\ \\

\dashrightarrow\:\: \sf  Height=15 + 8.65 \\  \\

\dashrightarrow\:\: \underline{ \boxed{ \sf  Height=23.65 \: m}} \\

\therefore \: \underline{\textsf{The height of the tower is \textbf{23.65 m}}}. \\

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