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Trigonometry Word problem with angles of elevation

  1. Jun 12, 2013 #1
    1. The problem statement, all variables and given/known data

    An Antenna that is 90 feet high is on top of a hill. From a point at the base of the hill, the angles of elevation to the top and bottom of the antenna are 28.5° and 25°, respectively. To the nearest whole number of feet, how high is the hill?

    A, 189 ft
    B, 213 ft
    C, 548 ft
    D, 623 ft
    E, 697 ft

    2. Relevant equations

    well I know Sinθ=o/h Cosθ=a/h Tanθ=o/a

    3. The attempt at a solution

    I don't know the length of the base so I couldn't figure it out like the textbook examples, but from angle 25° I figured the hypotenuse to the base of the antenna was 212.5, which would round to 213 so I know A and B can't be it.
     
    Last edited by a moderator: Jun 12, 2013
  2. jcsd
  3. Jun 12, 2013 #2

    Mark44

    Staff: Mentor

    I hope you have a drawing of the two triangles that represent this situation. You have the hypotenuse of the right triangle whose acute angle is 25°. That makes the other angle of this triangle 65°. You are given the length of the side opposite the 3.5° angle, and you should be able to figure out the obtuse angle of the upper triangle. Knowing two angles and a side, you should be able to use the Law of Sines to determine the hypotenuse of the upper triangle, and then get calculate the height of the hill.
     
  4. Jun 12, 2013 #3
    Where did you get 3.5° and where would the 28.5° angle go?
     
    Last edited: Jun 12, 2013
  5. Jun 12, 2013 #4

    LCKurtz

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    Gold Member

    Call the height of the hill ##h## and the distance from the base of the hill ##d##. Write the equations for the tangents of your two angles. Two equations in unknowns ##h## and ##d##. Eliminate the ##d## and solve.
     
  6. Jun 12, 2013 #5

    Mark44

    Staff: Mentor

    28.5° - 25° = 3.5°
    There are three triangles involved: the lower triangle (a right triangle with an acute angle of 25°); the upper triangle that has one angle of 3.5°; a large right triangle that contains the two previous triangles, and that has an acute angle of 28.5°.

    You have made a drawing, right?
     
  7. Jun 12, 2013 #6

    Mark44

    Staff: Mentor

    I like your strategy better than mine...
     
  8. Jun 12, 2013 #7
  9. Jun 12, 2013 #8

    Mark44

    Staff: Mentor

    That's angle of elevation, not evaluation. You had that in your title, and I fixed it.

    Your picture is wrong. You have the angle of elevation to the top of the antenna as 25°. It should be 28.5°. The angle of elevation to the bottom of the antenna is 25°.
     
  10. Jun 12, 2013 #9
    So where I had 25 degrees it should be 28.5, and 65 should be 61.5 degrees, correct?
     
  11. Jun 12, 2013 #10

    Mark44

    Staff: Mentor

    Yes.
     
  12. Jun 12, 2013 #11
    So where would the bottom angle of elevation go then?
     
  13. Jun 12, 2013 #12

    LCKurtz

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    Read your original post. It tells you.
     
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