Question

A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the

Answer 1

If both parts subtend equal angles at a point on the ground which is 25 metres away, then the height of the pole is 100√5 m or 223.60 m.

Step-by-step explanation:

Referring to the figure attached below,  

Let “BC" be the height of the verticle pole and let the point D divide the pole in such a way that  

BD : DC = 1 : 9 ........ [given]

The distance between the pole and the point where both the parts subtend equal angles, AB = 25 m

Let the equal angles subtended at point A by the two parts be ∠CAD =  ∠BAD = “θ”.

By using the Angle Bisector Theorem, we have

$\frac{BD}{DC} = \frac{AB}{AC}$

⇒ $\frac{1}{9} = \frac{25}{AC}$ ….. [substituting the given values]

⇒ AC = 225 m ….. (i)

Now, consider the right triangle ABC and apply the Pythagoras theorem,  

BC = √[AC² – AB²]

⇒ BC = √[225² – 25²]

⇒ BC = √[50000]

⇒ BC = √[10² * 10² * 5]

⇒ BC = 100√5 m or 223.60 m

Thus, the height of the vertical pole is 100√5 m or 223.60 m.

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