Two poles of equal heights are standing opposite each other on either side of the road, is 80 m wide. from a point between them on the road, the angles of elevation ofthe top of the poles are 60° and 30°, respectively. find the height of the poles and the distances of the point from the poles.​

Let AC and BD be the two poles of same height 'h' m.

Given AB = 80 m

Let AP = x m, therefore, PB = (80 - x) m

In ▶️APC,

In▶️ BPD,

Thus, the height of both the poles is m and the distances of the point from the poles are 60 m and 20 m.

Refer to the attachment for ur answer dear..

Step-by-step explanation:

As per above figure, AB = CD,

OB + OD = 80 m

Now,

In right ΔCDO,

tan 30° = CD/OD

1/√3 = CD/OD

CD = OD/√3 …..... (1)

Then,

In right ΔABO,

tan60° = AB/OB

√3 = AB/(80-OD)

AB = √3

AB = CD (Given)

√3(80-OD) = OD/√3 (Using equation (1))

3(80-OD) = OD

240 – 3 OD = OD

4 OD = 240

Putting the value of OD in equation (1)

CD = OD/√3

CD = 60/√3

CD=20√3 m

Also,

OB + OD =

OB = (80-60) m =

Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and 60 m respectively.

ab=Ed

ab=√3x=√3 x20

=20√3

Height of tower = 20√3m

" =20√3m

it is after the attachment first the attachment then this

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h = height of the poles

BE = x

EC = y

BC = 80m

In triangle ABE,

tan 60 = AB/BE

√3 = h/x

h = √3x--------(1)

In triangle DCE,

tan 30 = DC/EC

1/√3 = h/y

y/√3 = h--------(2)

x+y = 80

»y = 80-x--------(3)

Then eq(2) becomes

h = (80-x)/√3

(80-x)/√3 = √3x [from eq(1 and 2)]

80-x = 3x

80 = 3x+x

80 = 4x

x = 80/4 = 20m

substituting x = 20 in eq(3)

y = 80-x

= 80-20 = 60m

On substituting the value of x in eq(1) we get h = 20√3

= 34.64 approx.

Hope it will help you. . . . . . . . . .

answer:

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step-by-step explanation:

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