Let ABCD be a square, and let M and N be the midpoints of BC and CD respectively. Find sin<MAN.
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
stahp cheating
jk we encourage this
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)
YAY! GREAT!
He's not one of the founder, he IS the founder.
People, you realize that Richard Rusczyk is one of the founders of AoPs right.
Thanks a lot, sorry I am not too good at trigonometry
i like how its so easy to impersonate aops lol
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!