Question

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

Answer 1

Answer:

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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Answer 2

$\huge{\pink{\tt{\underline{Explanation:-}}}}$

______________________________

• Let AB and CD be the poles of equal height.

• O is the point between them from where the height of elevation taken.

• BD is the distance between the poles.

As per above figure, AB = CD,

OB + OD = 80 m

Now,

$\large\leadsto$In right ΔCDO,

$\large\leadsto$ tan 30° = CD/OD

$\large\leadsto$1/√3 = CD/OD

$\large\leadsto$ CD = OD/√3 …..... (1)

Then,

In right ΔABO,

$\large\leadsto$ tan60° = AB/OB

$\large\leadsto$ √3 = AB/(80-OD)

$\large\leadsto$ AB = √3

$\large\leadsto$ AB = CD (Given)

$\large\leadsto$ √3(80-OD) = OD/√3 (Using equation (1))

$\large\leadsto$ 3(80-OD) = OD

$\large\leadsto$240 – 3 OD = OD

$\large\leadsto$ 4 OD = 240

$</u><u>\</u><u>large</u><u>\leadsto$$\large{\pink{\tt{OD=60}}}$

Putting the value of OD in equation (1)

$\leadsto$ CD = OD/√3

$\leadsto$ CD = 60/√3

$\large\leadsto$CD=20√3 m

Also,

$\huge\leadsto$ OB + OD = $\star\boxed{\boxed{\blue{80 m}}}$

$\huge\leadsto$ OB = (80-60) m = $\star\boxed{\boxed{\blue{20m}}}$

$\huge\pink\star$ Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and 60 m respectively.