Question

the angle of elevation of the top of a vertical tower a point on the ground is 60 degrees from anothoe point 10 m ​

Answer 1

Answer:

let height of tower = x + 10m

let dist. between base of tower and obs. pt. = y

at first angle it makes with ground = 60 degree

tan(60) = sqrt{3} = (x + 10)/y   . . . . . . (eq 1)

next angle = 45 degree

pt is 10 m above the ground 

tan(45)= 1 = x/y    

:. x=y  . . . . . . . . . (eq. 2)

substituting value of y in eq. 1 

x + 10 = 1.732 x

:. x = 13.66 m 

Answer 2

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 :

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

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

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

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

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

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

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

⠀

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