急!急! F4 Properties Circles q9

19.
(a)
Let ∠DCB = y°

D is the mid-point of arc AB.
arc AD = arc DB
∠ACD = ∠BCD (equal arcs subtendequal ∠s)
∠ACD = y°

∠ACD + ∠BCD = ∠ACB
Then ∠ACB = 2y°

AB = AC (given)
Then ΔABC is an isos. Δ with ∠B = ∠ACB
∠B = 2y°

In ΔBCQ :
∠CQB + ∠B + ∠DCB = 180° (∠ sum of Δ)
102° + y° + 2y° = 180°
y = 26
Hence ∠DCB = 26°

(b)
In ΔACQ :
∠CQB = ∠ACD + ∠A (ext. ∠of Δ)
102° = 26° + ∠A
∠A = 76°

But ∠B = 2*26° = 52°
∠A ≠ ∠B
As only equal arcs subtend equal angles at circumference,
arc CB ≠ arc AC


20.
(a)
∠AOC = 2∠ABC (∠ at centre twice ∠ at circum.)
130° = 2∠ABC
∠ABC = 65°

(b)
BC = AC
Then ΔABC is an isos. Δ with ∠BAC = ∠ABC = 65°

In ΔABC :
∠ABC + ∠BAC + ∠BCA = 180° (∠ sum of Δ)
65° + 65° + ∠BCA = 180°
∠BCA = 50°

(c)
∠BOC = 2∠BAC (equal arcs subtendequal ∠s)
∠BOC = 2 * 65°
∠BOC = 130°

The circumference of the circle
= 5.2 ÷ (360/130) cm
= 14.4 cm


21.
(a)
arc AB : arc CD = 3 : 2
Since equal arcs subtend equal ∠s,
then ∠AOB : ∠COD = 3 : 2
∠AOB : 40° = 3 : 2
∠AOB = 60°

In ΔAOB :
OA = OB (radii of same circle)
Then ∠A = ∠B (equal sides to equal ∠s)
∠A + ∠B + ∠AOB = 180° (∠ sum of Δ)
∠A + ∠A + 60° = 180°
∠A = ∠B = ∠AOB = 60°
ABC is an equilateral Δ.
Hence, AB = 9 cm

(b)
∠AOB + ∠BOC + ∠COD = 180° (adj. ∠s on a st. line)
60° + ∠BOC + 40° = 180°
∠BOC = 80°

∠AOD = 180° (∠ on a st. line)

arc AB : arc BC : arc CD : arc DA
= ∠AOB : ∠BOC : ∠COD : ∠AOD
= 60° : 80° : 40° : 180°
= 3 : 4 : 2 : 9