Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Five angles whose sum is exactly ninety degrees?

Show that

π/2 = arcsin(12/37) + arcsin(57/185) + arcsin(3232/10457) + arcsin(7735/25033) + arcsin(385468067/1308850405)

exactly.

Update 2:

@Duke. Great solution! Also great picture! Thank you. Such terrible relationships are abundant with Pythagorean triangles and not hard to find (once you know how) with a little help from Maple. I'm so happy you are still on Y!A.

1 Answer

Relevance
  • Duke
    Lv 7
    6 years ago
    Favorite Answer

    To answer this question I needed some nuclear weaponry, sorry!

    Let θ₁, θ₂, θ₃, θ₄, θ₅ are the 5 angles listed in order

    (i.e. θ₁ = arcsin(12/37), etc.). All of them are acute angles in the following Pythagorean triangles:

    #1: (12, 35, 37); #2: (57, 176, 185); #3: (3232, 9945, 10457);

    #4: (7735, 23808, 25033); #5: (385468067, 1250801244, 1308850405)

    Closer look at the hypotenuses (denominators of your fractions) shows that 37 in #1 is prime and 185 = 5*37 in #2; 10457 in #3 and 25033 in #4 are prime, but 1308850405 = 5 * 10457 * 25033. That was the reason for me to try to add separately θ₁ + θ₂ and θ₃ + θ₄ + θ₅ using the relationships

    arcsin(x) + arcsin(y) = arcsin(x√(1 - y²) + y√(1 - x²))

    arcsin(x) + arcsin(y) + arcsin(z) = arcsin(x√(1 - y²)√(1 - z²) +

    + y√(1 - z²)√(1 - x²) + z√(1 - x²)√(1 - y²) - xyz)

    (valid because θ₁ + θ₂ < π/2, θ₃ + θ₄ + θ₅ < π/2)

    http://en.wikipedia.org/wiki/List_of_trigonometric...

    So θ₁ + θ₂ = arcsin((12*176 + 57*35)/(37*185)) = arcsin(3/5);

    much more efforts finally produced the following:

    θ₃ + θ₄ + θ₅ = arcsin((3232 * 23808 * 1250801244 + 7735 * 9945 * 1250801244 + 385468067 * 9945 * 23808 - 3232 * 7735 * 385468067) / (5 * 10457² * 25033²)) = ... = arcsin(4/5)!

    Then (θ₁ + θ₂) + (θ₃ + θ₄ + θ₅) = arcsin(3/5) + arcsin(4/5) = π/2 exactly, as the famous Egyptian Triangle (3, 4, 5) shows. So finally we are done.

    How on Earth and where do You dig such terrible relationships?

    P.S. You are mostly welcome, Rita! Yes, Pythagorean Triangles are a real Universe, but keeping abreast of mathematical software and updating personal knowledge becomes too much for an old man like me. Long Live the WWW!

    Attachment image
Still have questions? Get your answers by asking now.