Asked by Knights

Let ABCD be a square, and let M and N be the midpoints of BC and CD respectively. Find sin<MAN.
Answered by Reiny
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
Answered by Reiny
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)
Answered by Knights
Thanks a lot, sorry I am not too good at trigonometry
Answered by Writeacher
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!
Answered by AoPS
Stop cheating on your homework, it does not benefit you.
Answered by aops real one 100%
stahp cheating
jk we encourage this
Answered by Richard Rusczyk
YAY! GREAT!
Answered by Justin
People, you realize that Richard Rusczyk is one of the founders of AoPs right.
Answered by XP
He's not one of the founder, he IS the founder.
Answered by johnny v
i like how its so easy to impersonate aops lol
There are no AI answers yet. The ability to request AI answers is coming soon!