Question

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

Answer 1

Its NCERT question brdr!

Here is the solution;)

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

Answer 2

_____________________________

|............|.\P............/|X............................

|............|...\........../.|........................ ......

|........h.|.....\......./..|.h..............................

|............|.......\.../....|..................................

|............|__._..\/.__|.................................

|.........,.Q.......R....Y.....................................

|________________________________

/_PRQ=60°. /_XRY=30°

PQ=XY= height of the pole = h

QY=80m

.: QR=x RY=80-x

In ∆PQR

$tan30 = \frac{h}{x} \\ (tan30 = \frac{1}{ \sqrt{3} } ) \\ \frac{h}{x} = \frac{1}{ \sqrt{3} } \\ x = h \sqrt{3}$

x=h√3 -------------------(1)

In ∆XYR

$tan60 = \frac{h}{80 - x} \\ tan60 = \sqrt{3} \\ \frac{h}{80 - x} = \sqrt{3} \\ h = (80 - x) \sqrt{3}$

Put (1) in

h = (80-x)√3

h = (80- h√3)√3

h = 80√3 - 3h

h+3h = 80√3

4h = 80√3

h = 20√3

i.e. h = 34.64m

Height of pole = 34.64m