Page 132 - Mathematics_Form_3

Circles

̂
̂
Since, A P B = A Q C (angles subtended Exercise 5.4
by equal arcs on the circumference), it
implies that 1. If O is the centre of the circle, find
the value of x .
̂
A Q C = 40°
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(a)
̂
̂
̂
Similarly, A X C = Q A X + A Q X
(exterior angle of a ΔAQX ) 300º
Thus, O F
̂ ̂ ̂
Q A X = A X C − A Q X x

̂ ̂
= 70° − 40° ( A Q X = A P B)
= 30° D E
̂
Hence, Q A X = 30° .
̂
̂
But, A R B = Y R Q (vertically opposite (b) B
angles). It implies that x
̂
̂
̂
A R B = P A R + A P R (exterior angle of

ΔAPR ) 3x
Thus,
̂ ̂ O
Y R ̂ Q = P A R + A P R

̂
̂
= Q A X + A P B
A C
= 30° + 40°
= 70°
2. The angle subtended at a point
In ΔYRQ, it follows that
̂
̂
̂
Q Y R = 180° − (Y R Q + R Q Y ) ̂ on the circumference by an arc is
37 . Find the angle subtended by
o
=180° − (70° + 40° ) the same arc at the centre.
= 180° − 110° Mathematics for Secondary Schools
3. In a circle, an arc subtends
= 70°
o
̂ ̂ an angle of 120 at the centre.
But, Q Y R = Y R Q = 70°. Find the angle subtended at the
̂
Therefore, Y R Q = 70° . circumference by the same arc.
̂
̂
Since Q Y R and Y R Q are base angles of 4. In the following figure, O is the
an isosceles ∆ YRQ , then ‾ QY = ‾ QR .
centre of the circle.

Student\s Book Form Three 125

18/09/2025 09:59:39
MATHEMATIC F3 SB.indd 125
MATHEMATIC F3 SB.indd 125 18/09/2025 09:59:39