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

Answer 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

Answer 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)

Answer 3

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

Answer 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!

Answer 5

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

Answer 6

stahp cheating
jk we encourage this

Answer 7

YAY! GREAT!

Answer 8

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

Answer 9

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

Answer 10

i like how its so easy to impersonate aops lol