Math App of trig. in 3-D&(20s)

23.
(a)
In rt.Ð-ΔADF : DF² = AF² + AD² (Pythagoreantheorem)
DF² = (6² + 8²) cm²
DF = 10 cm

In rt.Ð-ΔABF : BF² = AF² + AB² (Pythagoreantheorem)
BF² = (6² + 8²) cm²
BF = 10 cm

In rt.Ð-ΔABD : BD² = AD² + AB² (Pythagoreantheorem)
BD² = (8² + 8²) cm²
BD² = 128 cm²
BD = 8√2 cm

In ΔFBD : cosÐDFB = (DF² + BF² - BD²) / (2 ´ DF ´ BF) (cosine law)
cosÐDFB = (10² + 10² - 128) / (2 ´ 10 ´ 10)
cosÐDFB = 0.36
ÐDFB = 68.9°
Angle between DF and BF = 68.9°


(b)
AM = (1/2)AC = (1/2)BD
Hence AM = (1/2) ´ 8√2 cm = 4√2 cm

In rt.Ð-ΔFAM :
tanÐAFM = AM / AF
tanÐAFM = 4√2 / 6
ÐAFM = 43.3°

Angle between FM and the plane EFGH
= 90° - 43.3°
= 46.7°


(c)
In rt.Ð-ΔFAM :
tanÐAMF = AF / AM
tanÐAMF = 6 / 4√2
ÐAMF = 46.7°

Angle between the planes BDF and BCD
= 180° - 46.7°
= 133.3°

2011-10-15 18:06:59 補充：
Alternative method for part (a) :

DF = 10 cm

In rt.∠-ΔABD :
BD² = (8² + 8²) cm² (Pythagorean theorem)
BD = 8√2 cm
DM = (1/2) x 8√2 cm = 4√2 cm

In rt.∠-ΔDMF :
sin∠DMF = 4√2 / 10
∠AFM = 34.45°

Angle between FM and the plane EFGH
= 2 x 34.45°
= 68.9° ...... (answer)