a vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it in which lower part is shorter than the upper part. If both parts subtend equal angles at a point on the ground, 15 metre away. Find the height of the pole.
Answers
If both parts subtend equal angles at a point on the ground which is 15 metres away, then the height of the pole is 134.16 m or 60√5 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 = 15 m
Let the the equal angles subtended at point A by the two parts be ∠CAD = ∠BAD = “θ”.
By using the Angle Bisector Theorem, we have
BD/DC = AB/AC
⇒ 1/9 = 15/AC ….. [substituting the given values]
⇒ AC = 135 m ….. (i)
Now, consider the right triangle ABC and apply the Pythagoras theorem,
BC = √[AC² – AB²]
⇒ BC = √[135² – 15²]
⇒ BC = √[15² {9² - 1}]
⇒ BC = √[15² * 80]
⇒ BC = 4*15 √5
⇒ BC = 60√5 m or 134.16 m
Thus, the height of the vertical pole is 60√5 m or 134.16 m.
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