I need some help on my math homework and I have no clue on how to go about it.?

Two Airplanes leave the same airport in the opposite directions. At 2:00P.M the angle of elevation from the aiport to the first plane is 26 degrees and to the 2nd plane is 17 degrees. The first plane has an elevation of 7km. The 2nd plane has an elevation of 8.2km find the line of sight distance between the 2 airplanes.

pls help :c

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  • 5 months ago

    Horizontal distance between two planes = hor. dist. of 1st plane + hor. dist. of 2nd plane

    hor. dist. = 7 x cos 26° + 8.2 x cos 17° = 14.1332 Km

    Vertical distance difference between two planes = 8.2 - 7 = 1.2 Km

    by Pythagorean Theorem

    sight distance = ( 14.332 ² + 1.2 ² ) ^ (1/2) = 14.3821 Km

    • Alan
      Lv 7
      5 months agoReport

      Elevation is 7 km and 8.2 km
      so the horizontal distance is not found using cosine
      7 is opposite the 26 degree angle
      cos= adj/hyp

  • 5 months ago

    Think of the y axis as elevation, and the positive x axis as the (ground) direction of the first plane. Then its coordinates at 2 p.m. are (7 cot(26), 7) and the coordinates of the 2nd plane are (-8.2 cot(17), 8.2). So the distance between the two airplanes is

    sqrt{[7 cot(26) + 8.2 cot(17)]^2 + (8.2 - 7)^2}

    = 41.19 km

  • Mike G
    Lv 7
    5 months ago

    Let the aircraft be a and b

    Angle between the aircraft = 180-26-17 = 137°

    Slant height of first aircraft = Sa

    Slant height of second aircraft = Sb

    Sa = 7/sin26 = 15.9682 km

    Sb = 8.2/sin17 = 28.0465 km

    Let x be the required distance

    Cosine Rule

    x^2 =

    15.9682^2 + 28.0465^2 - 2*15.9682*28.0465cos137

    x = 41.19 km

  • TomV
    Lv 7
    5 months ago

    The law of cosines can be used to calculate the line of sight (LOS) distance, c, between the two planes. You know the angle opposite that distance Θ = 180 - 26 - 17 = 137­°. What's missing is the LOS distance to each plane. But you are given all that's needed to find those distances.

    For the first plane, you know the altitude and elevation angle, 7 km and 17°. The LOS distance, a, can be found from the trigonometry of the situation:

    a = 7/sin17°

    For the second plane, the configuration is altitude 8.2 km and elevation angle 28­°. The LOS distance, b, to the second plane from the airport is:

    b = 8.2/sin28°

    That's all you need to apply the law of cosines to the problem:

    c² = a² + b² - 2ab cosΘ

    = (7/sin17)² + (8.2/sin28)² - 2(7/sin17)(8.2/sin28) cos 137

    = 573.225 + 305.077 + 611.680

    = 1489.983

    c = 38.6 km

    • Alan
      Lv 7
      5 months agoReport

      Math Typoes
      sin(26) changed to sin(28) why.
      You used 26 and 17 to find the 137
      then changed 26 to 28 (for no good reason.)

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  • 5 months ago

    Sketch the situation.....

    then

    if x is the horizontal distance from the airport for plane 1

    tan(26) = 7/x

    ==> x = 7/tan(26)

    if y is the distance from the airport for plane 2

    tan(17) = 8.2/y

    ==> y = 8.2/tan(17)

    line of sight (call it w) between the panes is thus

    (x+y)^2 + (8.2 - 7)^2 = w^2

    solve for w.....

    • Alan
      Lv 7
      5 months agoReport

      No, what you just said is exactly
      the same as the distance formula. The distance formula
      is totally based on the Pythagorean theorem
      (They aren't different.)
      distance = sqrt( (y - (-x))^2 + (8.2-7)^2 )
      distance = sqrt( (y+x)^2 + (8.2 -7)^2 )
      same thing as what you just said, not different.

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