Let ABCD be a square, and let M and N be the midpoints of BC and CD respectively. Find sin<MAN.

10 answers

  1. Make a sketch, letting each side of the whole square be 2 (could be anything, so why not something simple ?)

    let angle MAN = Ø , let angle BAM = angle DAN = y
    in triangle ABM
    AB = 2, BM = 1, then AM = √5 by Pythagoras
    and sin y = 1/√5

    at the right angle A
    x + 2y = 90
    x = 90 - 2y
    take the sine of both sides
    sinx = sin(90 - 2y)
    but sin(90-2y) = cos 2y by the complementary angle theorem
    and cos 2y = 1 - 2 sin^2 y

    so
    sinx = cos2y = 1 - 2(1/√5)^2
    = 1 - 2/5
    = 3/5 or .6

  2. don't know why I switched from Ø to x

    so at top, let angle MAN = x

    (I bet you could have figured that out yourself)

  3. Thanks a lot, sorry I am not too good at trigonometry

  4. Please do not post answers-only for other students' posts, especially those in which the other student has clearly not included any thinking on his/her own.

    Posting only answers (especially with no explanation) doesn't teach anyone anything ... except maybe how to cheat!

  5. Stop cheating on your homework, it does not benefit you.

  6. stahp cheating
    jk we encourage this

  7. YAY! GREAT!

  8. People, you realize that Richard Rusczyk is one of the founders of AoPs right.

  9. He's not one of the founder, he IS the founder.

  10. i like how its so easy to impersonate aops lol

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