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melisa1 [442]
3 years ago
10

Two poles of equal heights are standing opposite to each other on either side of a road, which is 100 meters wide .From a point

between them on the road, the angles of elevation of their tops are 30? and 60? .Find the position of the point and also, the height of the poles. ( use = 1.73)
Mathematics
2 answers:
ASHA 777 [7]3 years ago
8 0

Answer:

  • Distance to the pole from the vertex of the 60o angle = 25 m
  • Distance to the second pole from the vertex of the 30o angle = 75 m
  • Height of the pole = 43.3

Step-by-step explanation:

The point is somewhere on the road. From that point, two poles can be seen at different angles of elevation. One is 30 degrees, and another is 60 degrees.

Call the height of the poles y and the point on the road as x meters away from the base of 60 degree angle to the base pole

  • Tan(60) = y/x
  • Tan(30) = y/(100 - x)  Isolate y for both equations

  • x*tan(60) = y
  • (100 -x) * tan(30) = y  Since both boles are equal in height, equate the two equations.

(100 - x) * tan(30) = x*tan(60)

  • Using exact values (from the unit circle)
  • tan(30) = sqrt(3)/3
  • tan(60) = sqrt(3)
  • (100 - x) * sqrt(3)/3 = x*sqrt(3)     Divide both sides by sqrt(3)
  • (100 - x)/3 = x                              Multiply both sides by 3
  • 100 - x = 3x                                  Add x to both sides
  • 100 = 4x                                       Divide by 4
  • x = 25

The distance to the other pole = 100 - 25 = 75

  • y = x * sqrt(3)
  • y = 25 * 1.73
  • y = 43.3
Anuta_ua [19.1K]3 years ago
5 0
Let the point be 'P' whose distance from the left pole is 'x' m and from the right pole is (100 - x) m. Let the angle of elevation towards the left be 30° and towards the right be 60°. Let the heights of the poles be 'h'. Hence,
\tan( {30}^{0} )  \:  =  \:  \frac{h}{x}
h \:  =  \: x  \tan( {30}^{0} )
x \:  =  \: h \cot( {30}^{0} )
x \:  =  \: h \sqrt{3}
\tan( {60}^{0} )  \:  =  \:  \frac{h}{100 - x}
100 - x \:  =  \: h \cot(  {60}^{0}  )
100 - x \:  =  \:  \frac{1}{ \sqrt{3} } h
100 \sqrt{3 }   \:  -  \:  \sqrt{3} x \:  =  \: h
Substituting the value of x, we get,
100 \sqrt{3}  \:  -  \:  \sqrt{3} ( \sqrt{3} h) \:  =  \: h
100 \sqrt{3}  \:  -  \: 3h \:  =  \: h
4h \:  =  \: 100 \sqrt{3}
h \:  =  \: 25 \sqrt{ 3}
or,
h ≈ 25(1.73)
h ≈ 43.25 m
x \:  =  \:  \sqrt{3} h
x \:  =  \:  \sqrt{3} ( 25 \sqrt{3} )
x \:  =  \: 25(3)
x = 75 m.
Therefore,
h = 43.25 m, x = 75 m, and (100 - x) = 25 m.
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