S.5 Trigo 3-D

None of the options is correct.
The correct answer is 71.6 degree.

Detailed steps are here:
http://farm6.staticflickr.com/5160/6923142262_b6e6f5925e_b.jpg

Firstly , let K be the mid-point of EM and
O be a point on MH
By repeatly Pythagoras theorem,we can get
AM=10√2 cm ； AE=4√29 cm；EM=2√34 cm；KM=√34 cm.
cosine theorem:
AE^2=AM^2+EM^2-2(AM)(EM)cos∠EMA
464=200+136-40√68 cos∠EMA
cos∠EMA=[(464-336)/ -80√17]
∠EMA=cos^-1 ((336-464)/80√17)
∠EMA=cos^-1 (-128/80√17)
-----------------------------------------------
cosine theorem:
AK^2=KM^2+AM^2-2(KM)(AM)cos∠EMA
AK^2=34+200-20√68 coscos^-1 (-128/80√17)
AK^2=234-40√17 * -128 / 80√17
AK^2=234+128/2
AK=√298
------------------------------------------------
By intercept theorem:
∵MK=EK and KO//EH
∴MO=HO
MO=1/2 *10=5 cm
∴CO=15cm
-----------------------------------------------
∠KAC=sin^-1(CO/AK)
=sin^-1 (15/√298)
=60.3(corr.to 1 d.p.)//
-----------------------------------------------
Ans:C
p.s.:因為AEM不是平衡於ABCD,要用AK作為AEM的中間軸去計算兩面之間的。