唔識得做！！！！！！（１５Ｘ２）

1.
angle (ZPQ) = angle (ZXY) = 32 (corr. angles, PQ//XY)
angle (PQY) = angle (ZPQ) + angle (PZQ) (ext. angle of triangle)
b = 32 + 90 = 122


2.
There are not enough points for me to do.
I just name them. HG//EF//CD [angle (HGE) = 45, angle (GHA) = b]

angle (HGB) = angle (CDB) = 110 (corr. angles, HG//CD)
angle (EGB) = angle (HGB) - angle (HGE) = 110 - 45 = 65
angle (GDA) = angle (EGB) + angle (GEA)
b = 65 + 90 = 155


3.
Draw 3 vertical lines at 3 different angles, named from left to right, PQ, RS, TD,
where angle (AQR) = 132, angle (QRV) = c and angle (RDE) = 136

As PQ and TD are vertical lines, angle (AQP) = 90 and angle (TDE) = 90
angle (PQR) = angle (AQR) - angle (AQP) = 132 - 90 = 42
angle (RDT) = angle (RDE) - angle (TDE) = 136 - 90 = 46
angle (QRS) = angle (PQR) = 42 (alt. angle, PQ//RS)
angle (SRD) = angle (RDT) = 46 (alt. angle, RS//TD)
angle (QRD) = angle (QRS) + angle (SRD) = 42 + 46 = 88


Sorry for not providing the pictures for the solutions.
If you have any problem about reading this, please feel free to send me an-email.

2007-01-20 22:20:52 補充：
答案跟先前一樣, 如果看不明白, 可隨時問我. ^_^

∠ZPQ=∠ZXY=32
b=∠ZPQ+90
b=122

∠EGB=110-45=65
b=∠EGB+90=155

想像一條穿過角C果點又平行於DE既線
132+(136-c)=180
c=88

2007-01-20 20:41:45 補充：
第一題個答案好明顯大過90度點解咁多人計到58度既?

我只睇得明你第一題:
PQ// XY,

2007-01-20 20:36:24 補充：
剛才SHOW唔到~~~ PQ// XY, 角ZXY=角ZPQ= 32' AND '.' 角ZQP⊥,=90' △ZQP=角ZQP 角ZPQ 角PZQ= 180' 180' = 角ZPQ 90' 32' 角ZPQ= 58'唔好理個"大"字....唔知做乜...SHOW 唔到後面的數

2007-01-21 02:05:25 補充：
b= 180-58'=132'......sor=.=:"