Question

The angle of elevation of the top of a pillar from point on the ground is 15° on walking 100 m towards the tower, the angle of elevation is found to be 30.Calculate the height of the tower (where tan 15 =2-13).


please solve it correctly ...........​

Answer 1

Answer :

Height of tower = 50 m

Solution :

Let, height of the tower be 'h'.

From the figure,

In ∆ABD,

θ = 30°

$tan \: 30° = \frac{h}{x}$

$\frac{1}{\sqrt{3}} = \frac{h}{x}$

x = h√3 ----- eq 1

In ∆ABC,

$tan \: 15° = \frac{h}{x + 100}$

$2-\sqrt{3} = \frac{h}{x + 100}$

$x = \frac{h}{2 - \sqrt{3}}-100$ ------ eq 2

From equation 1 and 2,

$h\sqrt{3} = \frac{h}{2 - \sqrt{3}}-100$

$h\sqrt{3} + 100 = \frac{h}{2 - \sqrt{3}}$

$(h\sqrt{3} + 100)(2 - \sqrt{3}) = h$

$2h\sqrt{3}- 3h + 200 - 100\sqrt{3} = h$

$200 - 100\sqrt{3} = 4h - 2h\sqrt{3}$

$100 - 50\sqrt{3} = 2h - h\sqrt{3}$

$50(2 - \sqrt{3} )= h(2 - \sqrt{3})$

h = 50

∴ Height of tower = 50 m

Answer 2

✿ Solution :- ✿

Let the height of tower be H & Base of ∆₁ be x

By observing the given figure ,

tan30° = H/x

Substituting tan30° = 1/√3

⇒ 1/√3 = H/x

By cross - multiplying ,

⇒ x = H√3 …eq(1)

Now , ∆₂

tan15° = H/x + 100m

Substituting , tan15° = 2 - √3 [ ∵ Given ]

⇒ 2 - √3 = H/x + 100m

⇒ 2 - √3 = H/[ H√3 + 100 ] [ ∵ eq(1) ]

By cross - multiplying ,

⇒ H = ( 2 - √3 ) [ H√3 + 100 ]

⇒ H = 2H√3 + 200 - H(√3)(√3) - 100√3

⇒ H = 2H√3 + 200 - 3H - 100√3

⇒ H + 3H - 2H√3 = 200 - 100√3

⇒ 4H - 2H√3 = 200 - 100√3

⇒ H( 4 - 2√3 ) = 200 - 100√3

⇒ H = 200 - 100√3/( 4 - 2√3 )

⇒ H = 100( 2 - √3 )/2 ( 2 - √3 )

⇒ H = 100/2

⇒ Height of tower = 50 m

Hence , height of tower = 50m