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Two poles of equal heights are standing opposite each other on either side of the road which is 85 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively.
Two poles of equal heights are standing opposite each other on either side of the road which is 85 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles. (use $\sqrt{3}$ = 1.73)
This Question has 1 answers.
Let the height of each pole be $h$ meters.
Let the point on the road be at distances $x$ meters and $(85 - x)$ meters from the two poles.
Using the tangent formula:
For the pole at distance $x$:
$ \tan 60^\circ = \frac{h}{x} $
Since $ \tan 60^\circ = \sqrt{3} $, we get:
$ \sqrt{3} = \frac{h}{x} $
$ h = \sqrt{3} x $ \quad (Equation 1)
For the pole at distance $(85 - x)$:
$ \tan 30^\circ = \frac{h}{85 - x} $
Since $ \tan 30^\circ = \frac{1}{\sqrt{3}} $, we get:
$ \frac{1}{\sqrt{3}} = \frac{h}{85 - x} $
$ h = \frac{(85 - x)}{\sqrt{3}} $ \quad (Equation 2)
Equating both expressions for $h$:
$ \sqrt{3} x = \frac{85 - x}{\sqrt{3}} $
Multiplying both sides by $\sqrt{3}$:
$ 3x = 85 - x $
$ 3x + x = 85 $
$ 4x = 85 $
$ x = \frac{85}{4} = 21.25 $ meters
Substituting $x$ in Equation 1:
$ h = \sqrt{3} \times 21.25 $
Using $ \sqrt{3} \approx 1.732 $:
$ h = 1.732 \times 21.25 = 36.8 $ meters
The height of each pole is $36.8$ meters.
The distances of the point from the poles are $21.25$ meters and $63.75$ meters.
Let the point on the road be at distances $x$ meters and $(85 - x)$ meters from the two poles.
Using the tangent formula:
For the pole at distance $x$:
$ \tan 60^\circ = \frac{h}{x} $
Since $ \tan 60^\circ = \sqrt{3} $, we get:
$ \sqrt{3} = \frac{h}{x} $
$ h = \sqrt{3} x $ \quad (Equation 1)
For the pole at distance $(85 - x)$:
$ \tan 30^\circ = \frac{h}{85 - x} $
Since $ \tan 30^\circ = \frac{1}{\sqrt{3}} $, we get:
$ \frac{1}{\sqrt{3}} = \frac{h}{85 - x} $
$ h = \frac{(85 - x)}{\sqrt{3}} $ \quad (Equation 2)
Equating both expressions for $h$:
$ \sqrt{3} x = \frac{85 - x}{\sqrt{3}} $
Multiplying both sides by $\sqrt{3}$:
$ 3x = 85 - x $
$ 3x + x = 85 $
$ 4x = 85 $
$ x = \frac{85}{4} = 21.25 $ meters
Substituting $x$ in Equation 1:
$ h = \sqrt{3} \times 21.25 $
Using $ \sqrt{3} \approx 1.732 $:
$ h = 1.732 \times 21.25 = 36.8 $ meters
The height of each pole is $36.8$ meters.
The distances of the point from the poles are $21.25$ meters and $63.75$ meters.
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