Question

Two poles are erected on either bank of a river just opposite to each other. One pole is 40m high. From the top and foot of this pole, the angles of elevation of the top of the other pole are 30 degree and 60 degree respectively. Find the height of the other pole and width of the river.

Answer 1

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Answer 2

Answer:

The width of river is 20√3 m and height of another tower is 60 m.

Step-by-step explanation:

Refer the attached figure

Height of first pole = AB = 40 m

CE is the another pole

From the top and foot of AB pole, the angles of elevation of the top of the other pole are 30 degree and 60 degree respectively.i.e. ∠EAD = 30° and ∠EBC = 60°

Now we are supposed to find the height of another tower i.e. CE and width of river i.e. BC = AD

AB = DC = 40 m

Let ED be x

So, EC = ED +DC = x+40

In ΔEAD

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

$tan 30^{\circ} = \frac{ED}{AD}$

$\frac{1}{\sqrt{3}}= \frac{x}{AD}$

$AD= \sqrt{3}x$   --1

In ΔEBC

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

$tan 60^{\circ} = \frac{EC}{BC}$

$\sqrt{3}= \frac{x+40}{BC}$

$BC= \frac{x+40}{\sqrt{3}}$   -2

Since BC = AD

Equate  1 and 2

$\frac{x+40}{\sqrt{3}}= \sqrt{3}x$

$x+40=3x$

$40=2x$

$20=x$

So, EC = ED+DC =x+40=20+40 =60 m

$AD= \sqrt{3}x=2-\sqrt{3}$

So, the width of river is 20√3 m and height of another tower is 60 m.