Page 138 - Mathematics_Form_3

Circles

Solution Solution
̂
̂
From the given figure, construct lines In the figure, A B C = A O C (angle at
1 __
2
PM an RM as shown in the following the centre). It implies that,
figure. ̂ 1 __
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Q A B C = × 110°
2

x R = 55°
y But, ∠ABC + ∠ADC = 180° (opposite
P angles in a cylic quadrilateral). Thus,
O
∠ADC = 180° − ∠ABC

= 180° − 55°
= 125° .

M Therefore, ∠ADC is 125° .
From the figure, it implies that
̂ ̂
P O R = 2 P M R (Angle at the centre).
̂ ̂ 1 __ Example 5.11
But, P O R = y. Hence, P M R = y.
2 In the following diagram, find the
1 __
Thus, y + x = 180° (Opposite angles ˆ ˆ
2 measures of STU and SVU.
in a cyclic quadrilateral).
T
Therefore, y + 2x = 360°.

S
Example 5.10
In the following diagram, O is the 85°
centre of a circle. If ∠AOC = 110° , 40° U
R V
find ∠ADC.
B Solution
ˆ
In RTU,∆ 85°+ 40°+ RTU 180= °
(sum of angles in a triangle) Mathematics for Secondary Schools

ˆ
Thus, RTU 180= °− 125°
O
55= °
ˆ
110º But, RTU STU.= ˆ
A ˆ
⇒ STU 55= °
C SVU STU 180+ ˆ ˆ = ° (opposite angles
D in cyclic quadrilateral)

Student\s Book Form Three 131

18/09/2025 09:59:42
MATHEMATIC F3 SB.indd 131 18/09/2025 09:59:42
MATHEMATIC F3 SB.indd 131