Page 131 - Mathematics_Form_3

Circles

̂
ˆ
Thus, AOB 2OXA,= ˆ . (b) D O E + 200° = 360° (sum of
angles in a circle)
ˆ
ˆ
But OXA = AXB . D O E = 360° − 200° ,
̂
ˆ
Therefore, AOB 2AXB.= ˆ = 160° .
̂
FOR ONLINE READING ONLY
Therefore, D O E is 160°.
̂
̂
1 __
Example 5.5 (c) D F E = D O E (angle at the
2
centre)
Use the following figures to find: ̂ 1 __
̂ D F E = × 160°
2
(a) A O B
̂ = 80°
(b) D O E
Therefore, x = 80° .
(c) The value of x.
Example 5.6
C
In the following figure, O is the centre
50º of a circle with arc AB = arc AC.
The reflex angle AOB = 280° and
̂
̂
A X C = 70° . Calculate Y R Q and
O hence, show that ‾ QY = ‾ QR
B P

C X Y Q
A 70º
280º R

O

Mathematics for Secondary Schools Solution E x F Solution ̂ ̂ ̂ B
200º
O
A

̂
A O B + 280° = 360° (sum of angles in

a circle). Thus, A O B = 80°
D
But A O B = 2A P B (angle at centre
is twice the angle subtended at the
̂
(a) 2A C B = A O B (angle at the
̂
̂
⇒A P B = A O B (angles at centre)
1 __
̂ ̂ centre). circumference by the same arc).
2
2(50° ) = A O B = × 80°
1 __
̂
Therefore, A O B = 100° . 2
̂
Hence, A P B = 40° .
124 Student\s Book Form Three

18/09/2025 09:59:38
MATHEMATIC F3 SB.indd 124
MATHEMATIC F3 SB.indd 124 18/09/2025 09:59:38