The angle of elevation of the top of a vertical pole as seen from a point 10 meters away from the pole is double its angle of elevation as seen from a point 70 meters from the pole. find the height of the pole above the level of the observer's eyes.

 59.2 m

Step-by-step explanation:

The tangent of the angle of elevation will be the ratio of pole height to distance from the pole:

  tan(2α) = h/10

  tan(α) = h/70

The double-angle formula for tangents tells us ...

  tan(2α) = 2tan(α)/(1 -tan(α)²)

Multiplying by the denominator and substituting from above, we get ...

  (1 -(h/70)²)(h/10) = 2(h/70)

  7(1 -(h/70)²) = 2 . . . . . . . . multiply by 70/h

  1 - 2/7 = (h/70)² . . . . . . . . divide by 7, subtract 2/7-(h/70)²; next: square root

  h = 70√(5/7) ≈ 59.2 . . . . meters

The height of the top of the pole is about 59.2 meters above the observer.

answer:

at time x=0 we have 1000, at time x=1 we have 2000, at time x=2 we have 4000.  

so that's an initial value of 1000 and a ratio of 2,

y = 1000 (2^x)

hope this !