(15分)有冇人知呢條數點計

圖片參考：http://i.imagehost.org/0149/iecf71rd.gif

1)ㄥADB = 180 - 10 - 70 - 60 = 40°
ㄥAEB = 180 - 70 - 60 - 20 = 30°

ㄥBED = x + 30
ㄥBDE = 180 - 20 - (x + 30) = 130 - x

In △ABD , by sine law :

sin (10+70) / sin 40 = BD / AB

BD = (sin 80 / sin 40) AB......(1)

In △ABE , by sine law :

sin 70 / sin 30 = BE / AB

BE = (sin 70 / sin 30) AB......(2)

In △BDE , by sine law :

sin (x + 30) / sin (130 - x) = BD / BE , by (1) & (2) :

= (sin 80 sin 30)AB / [(sin 40 sin 70)AB]

= (sin 80 sin 30) / (sin 40 sin 70)

= (2sin 40 cos 40 sin 30) / (sin 40 sin 70)

= cos 40 / sin 70

= sin 50 / sin 70

We have :
sin (x + 30) sin 70 = sin (130 - x) sin 50

sin (x + 30) sin 70 = sin (x + 50) sin 50

(-1/2)[cos(x+30+70) - cos(x+30-70)] = (-1/2)[cos(x+50+50) - cos(x+50-50)]

cos(x + 100) - cos(x - 40) = cos(x + 100) - cos x

cos(x - 40) = cos x

cos(40 - x) = cos x

40 - x = x

x = 20°

2)ㄥADB = 180 - 50 - 60 - 24 = 50°
ㄥAEB = 180 - 60 - 50 - 30 = 40°

ㄥADE = 50 + (180 - 30 - 40 - x) = 160 - x

In △ABD is a issos.△ since ㄥB = ㄥD , we have AD = AB......(1)

In △ABE , by sine law :

sin(50 + 30) / sin 40 = AE / AB

AE = (sin 80 / sin 40) AB......(2)

In △ADE , by sine law :

sin x / sin (160 - x) = AD / AE , by (1) & (2) :

= AB / [(sin 80 / sin 40) AB]

= sin 40 / sin 80

= sin 40 / (2sin40 cos40)

= 1 / 2cos40

= (1/2) / cos40

= sin30 / cos40

= sin30 / sin50

We have :
sin (160 - x) sin 30 = (sin x) sin 50

sin (x + 20) sin 30 = (sinx) sin 50

(-1/2)[cos(x+20+30) - cos(x+20-30)] = (-1/2)[cos(x+50) - cos(x-50)]

cos(x + 50) - cos(x - 10) = cos(x + 50) - cos(x - 50)

cos(x - 10) = cos(x - 50)

cos(10 - x) = cos(x - 50)

10 - x = x - 50

2x = 60

x = 30°