Page 145 - Mathematics_Form_3

Circles

̂
Example 5.16 A D B = 180° – 117.25° .
= 62 . 75° .
In the following figure, ‾ BD is a Therefore, A D B = 62 . 75° .
̂
diameter of a circle whose centre is at
ˆ
ˆ
̂
ˆ
O. If B A C = 62 . 5° and AOD 54.5 ,= ° (d) ACB = ADB (angles in same
FOR ONLINE READING ONLY
find the values of: segment). Thus,
̂
̂
̂
(a) C A D (b) A B D A D B = 62 . 75° (calculated)

̂
Therefore, A C B = 62 . 75°.
̂
̂
(c) A D B (d) A C B
B Exercise 5.8
1. If ‾ AB is a diameter of a circle with
centre O and C is a point on the
̂
circle, show that A C B = 90° .
O 2. A circle with centre O has a
62.5º 54.5º diameter ‾ AB . If C is a point on

A C the circle such that ‾ BC = 2 ‾ AC ,
find the angle subtended by the arc
BC on the circumference.
D 3. A quadrilateral is inscribed in a
Solution circle where one of its diagonals is
(a) From the figure, it follows that the diameter. Given that one angle
̂
B A D = 90° (angle in a semi is 70°, find the measures of the
circle). remaining angles.

Thus, ̂ ̂ ̂ 4. In the following figure, if O is the
̂
centre of the circle, find:
Mathematics for Secondary Schools (b) A B D = A O D (angle at centre). (c) J L M (d) L O M
C A D + B A C = 90°
̂
̂
⇒ C A D + 62 . 5° = 90°
(b) K O L
(a) L M K
⇒ C A D = 90° − 62 . 5°
̂
̂

= 27 . 5°
̂

Therefore, C A D = 27 . 5° .
J
M
̂
̂
1 _
2
29º

̂
1 __
⇒ A B D = × 54 . 5°
2
= 27 . 25°
̂
Therefore, A B D = 27 . 25° . O
̂ ̂ ̂
(c) A D B + B A D + A B D = 180° K L

̂
A D B + 90° + 27 . 25° = 180°
138 Student\s Book Form Three
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MATHEMATIC F3 SB.indd 138 18/09/2025 09:59:45
MATHEMATIC F3 SB.indd 138