Question

the height of a building is 11 metre and the height of a pole is 8 metre watching from the midpoint of the line segment joining their base angle of elevation of the top of the pole is a and then of the top of the building is beta then what is dash building true​

Answer 1

Step-by-step explanation:

Height of pole AD = 15\sqrt{2}15

2

Height of pole EB= 30\sqrt{2}30

2

Step-by-step explanation:

Refer the attached figure

The distance between two vertical poles is 60 m i.e. AB = 60 m

The height of one of the pole is double the other.

Let the height of Pole AD be x

So, height of other pole BE = 2x

The angles of elevation of the top of the poles from the mid point of the line segment joining their feet are complementary.

⇒∠DCA+∠ECB=90°

Let ∠DCA be y

then

∠ECB=90°-y

Since we are given that angle of elevation was measured from the midpoint of AB

AC +CB=60

2AC = 60

AC =30

⇒AC=CB =30 m

In ΔADC

tan \theta =\frac {Perpendicular}{Base}tanθ=

Base

Perpendicular

\text{tan y}=\frac {AD}{AC}tan y=

AC

AD

\text{tan y }=\frac {x}{30}tan y =

30

x

---a

In ΔBEC

tan \theta =\frac {Perpendicular}{Base}tanθ=

Base

Perpendicular

tan (90-y)=\frac {EB}{CB}tan(90−y)=

CB

EB

tan (90-y) =\frac {2x}{30}tan(90−y)=

30

2x

Property: tan(90-\theta)^{\circ}= cot\thetatan(90−θ)

∘

=cotθ

So, \text{cot y} =\frac {2x}{30}cot y=

30

2x

Property Tan\theta = \frac{1}{Cot\theta}Tanθ=

Cotθ

1

\text{tan y} =\frac {30}{2x}tan y=

2x

30

--b

Equate a and b

\frac {x}{30}=\frac {30}{2x}

30

x

=

2x

30

\frac {x}{30}=\frac {15}{x}

30

x

=

x

15

x^2= 30 \times 15x

2

=30×15

x^2= 450x

2

=450

x=\sqrt{450}x=

450

x=15\sqrt{2}x=15

2

Thus the height of pole AD = 15\sqrt{2}15

2

Height of pole EB = 2x = 30\sqrt{2}30

2