Question

The angle of elevation of the top of a vertical tower from a point on the ground is 60 degree. From another point 10 m vertically above the first it's angle of elevation is 45 degree. Find the height of the tower ​

Answer 1

Given :-

◉ Angle of elevation of the top of a vertical tower from a point on the ground is 60°.

◉ 10m vertically above the first point, the angle elevation is 45°.

To Find :-

◉ Height of the tower

Solution :-

Please refer to the attachment for the diagram.

In the diagram, the height of the tower is 10 + x

Because we let AE = x

Now, In ∆ABC,

⇒ tan 60° = height of tower / BC

⇒ √3 BC = 10 + x [∵ tan 60° = √3 ]

⇒ BC = (10 + x) / √3 ...(1)

Similarly, In ∆ADE,

⇒ tan 45° = AE / DE

but, DE = BC, because DE and BC are opposite sides of a rectangle , and In a rectangle the opposite sides are equal and parallel as well.

⇒ 1 = x / (10 + x)/√3 [∵ tan 45° = 1 ]

⇒ 1 = √3 x / (10 + x)

⇒ 10 + x = √3x

⇒ 10 = √3x - x

⇒ 10 = x(√3 - 1)

⇒ x = 10 / (√3 - 1)

So, The height of the tower was 10 + x, substituting value of x

⇒ Height = 10 + 10/(√3 - 1)

⇒ Height = (10√3 - 10 + 10) / (√3 - 1)

⇒ Height = 10√3 / (√3 - 1)

Rationalising the denominator,

⇒ Height = 15 + 5√3

If we put √3 = 1.73

⇒ Height = 15 + 5×1.73

⇒ Height = 15 + 8.65

⇒ Height = 23.65

So, If we put √3 = 1.73

• Height of tower = 23.65 m

While if we dont,

• Height of tower = 5(3 + √3) m

Answer 2

★ Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\linethickness{0.4mm}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){4.4}}\put(8,3){\line(3,0){3}}\put(11.1,2){\sf 10 m}\put(7.3,2){\sf 10 m}\put(11.1,4){\sf x m}\qbezier(8,1)(10.7,5)(11,5.4)\qbezier(8,3)(10.5,5)(11,5.4)\qbezier(8.65,1)(8.7,1.4)(8.4,1.6)\put(8.7,1.2){\sf60^{\circ} }\qbezier(8.5,3.4)(8.8,3.2)(8.7,3)\put(8.8,3.2){\sf45^{\circ} }\put(7.7,3){\large{A}}\put(7.7,1){\large{B}}\put(11.1,1){\large{C}}\put(11.1,3){\large{D}}\put(11.1,5.3){\large{E}}\end{picture}$

______________________

Step-by-step explanation:

• 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,

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

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

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

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

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

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

_______________________

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

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

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

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

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

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

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

_______________________

★ Height of the tower will be :

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

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

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

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

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

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

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

⠀

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