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.

angle addition postulate

2x=56

step-by-step explanation:

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068