Question

The horizontal distance between two poles is 15 m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30°. If the height of the second pole is 24m, find the height of the first pole.

Answer 1

The height of the first pole is 15.34 m

Explanation:

Given that the horizontal distance between two poles is 15 m.

The angle of depression of the top of the first pole as seen from the top of the second pole is 30°.

If the height of the second pole is 24 m.

To find: The height of the first pole.

Let the poles be AB and CD where CD = 24 m

The angle of depression of the top of the first pole is 30°.

Thus, we have,

$\angle {CAE}=30^{\circ}$ and $B D=15 \ {m}$

Let the height of the pole AB be h.

From the figure, it is obvious that, $A E=B D=15$ and $A B=E D=h$

Thus, $\ {CE}=\ {CD}-\ {ED}$

$CE=24-h$

Let us consider the triangle ACL

We have,

$\tan 30^{\circ}=\frac{C E}{A E}$

      $\frac{1}{\sqrt{3} } =\frac{24-h}{15}$

Cross multiplying, we get,

 $\frac{15}{\sqrt{3}}=24-h$

$5 \sqrt{3}=24-h$

    $h=24-5 \sqrt{3}$

    $h=24-(5 \times 1.732)$

    $h=15.34$

Thus, the height of the pole AB is 15.34 m

Learn more:

(1) The horizontal distance between two poles is 15m. the angle of depression of the top of first pole as seen from thr top of second pole is 30 degree. if the pole is 24m , find the height of the first pole.

brainly.in/question/787615

(2) The horizontal distance between two Pole is 15 M the angle of depression of the top of the first policies from the top of second Pole is 30 degree if the height of the second phone is 24 M find the height of the first pole .

brainly.in/question/8582755

Answer 2

Step-by-step explanation:

Given that:

∠APB=60

∘

,∠CPD=30

∘

,AC=80m

To find:

The height of the pole=AB=CD=?

Solution:

Let AB and CD be the two poles of equal height and P be the point on the road between the poles.

In △APB,

tan60

∘

=

AP

AB

or, AP=AB×

tan60

∘

1

or, AP=

3

AB

−−−−−−−(i)

In △PCD,

tan30

∘

=

CP

CD

or, CP=CD×

tan30

∘

1

or, CP=

3

CD=

3

AB ∵AB=CD −−−−−−−(ii)

Adding eqn. (i) and eqn. (ii) we get,

AP+CP=

3

AB

+AB

3

or, AC=AB(

3

+

3

1

)

or, 80m=4

3

AB

or, AB=20

3

m

Therefore, height of the pole=20

3

m=34.64m