今日系青年好努力咁做本作業...但系做到e課後面d tangent就



又要向高人求助下了

答住一題先:
10a. Let OB intersects circumference 果點做E
因為OC=OE (radius)
所以 角OEC = (180-x)/2
角ACD = 角OEC = (180-x)/2 ( angles in alt. segment)

10b. BC=AC 所以 角CBA =角CAB
角CBA = 90-x
2(90-x) = (180-x)/2 (ext. angle of triangle)
x = 60度

11a. Length of AD
= root ( 12^2 + 12^2 )
= 12root2 cm

11b. consider triangle BCE
BE^2 + BC^2 = CE^2
8BE^2 + 8BC^2 = 8CE^2
8CD^2 + 8BC^2 = 5AD^2
8 (12)^2 + 8BC^2 = 5(12root2)^2
BC=6 cm
then AB=AD-BC
= 12-6
= 6cm

15a. BR=10cm CR=5cm
BC=10+5=15cm
CT=5cm
then BT= root (15^2 - 5^2) = 10root2 cm
Diameter= 10root2 cm

15b. AT=8cm
Area= (10+5+5+8) x 10root2 除2
= 140root2 cm^2

16. construct CM, such that CM 垂直於 AE
AM=10-6=4cm
CM= root(4^2 + 16^2) = 4root17 cm
Area = (10+6)x 4root 17 除2
= 32root 17 cm^2

17. CQ=AB=8cm
CP=root(10^2 - 8^2) = 6cm
AC+AP=6
besides, AP-AC=2
solve simultaneously, AC=2cm, AP=4cm
Radius of larger one = 4cm
Radius of smaller one = 2cm

22. 角HGD = 90-17 = 73度
角HDG = 180-73x2 = 34度 其一
角FGC=90-(90-66) = 66度
角FCG = 180-66x2 = 48度 其二
角AHE = 17度 (angles in alt. segment)
角HAE = 180-17x2 = 146度 其三
角BFE = 90-66 = 24度 (angles in alt. segment)
角EBF = 180-24x2 = 132度 其四

20. 角BAC=x (angle in the same segment)
角DAC=角BAC=x (eq.arcs,eq.angles)
角DAB=x+x=2x
角DBE=角DAB=2x (angles in alt.segment)
x+2x = 123 (ext. angle of triangle)
x = 41度

21a. 角DEB=63 (angle in alt. segment)
角EDB = 63 (base angles, isos. triangle)
角EBD = 180-63x2
= 54度

21b. Let 角BCA be x
角BAF= 角ABF = x (angle in alt. segment)
x+x = 156度 (ext. angle of triangle)
x = 78
角BCA=78度

30a. Let 角ABC be 2x
角ACB=x (arcs prop. to angles at circum.)
角DBC = 2x/2 = x (因為DB bisects angle ABC)
角DBC=角DCB=x
BD=DC (sides opp. eq. angles)
BDC is an isos. triangle

30b. CA=CB
角CBA=角CAB
then 角EAC = 角DBC
角ACE=角BCD (common angle)
CA=CB
then ACE is congruent to BCD (AAS)

GE2449~AV36 :答住一題先:
10a. Let OB intersects circumference 果點做E
因為OC=OE (radius)
所以 角OEC = (180-x)/2
角ACD = 角OEC = (180-x)/2 ( angles in alt. segment)

10b. BC=AC 所以 角CBA =角CAB
角CBA = 90-x
2(90-x) = (180-x)/2 (ext. angle of triangle)
x = 60度

本人不加point E的做法:

10a.
OA=OC (radii)
∠OAC=∠OCA (base ∠s, isos.△)
∠OAC+∠OCA=x (ext. ∠ of △)
2∠OAC=x
∠OCA=x/2
∠ACD+∠OCA+90°=180° (adi. ∠s on st. line)
∠ACD=90°-x/2
∠ACD=(180°-x)/2

10b.
∠OCA=x/2 [from (a)]
BC=AC (given)
∠OBC=∠OAC (base ∠s, isos.△)
∠ACD=∠OBC+∠OAC (ext. ∠ of △)
(180°-x)/2=2∠OAC
(180°-x)/2=2(x/2)
2x=180°-x
3x=180°
x=60°

睇完上面啲sol再對翻個圖就明了..
發覺成日睇唔倒d angle in alt. segment...
同埋tangent properties入面既兩隻角相等..搞到miss左d條件
Thanks at all.