Page 144 - Mathematics_Form_3

Circles

Example 5.14
C
In the following figure, ‾ PQ is a 21 cm
diameter of a circle and R is a point
on the circumference. If ∠QPR = 52° , O B
FOR ONLINE READING ONLY
find ∠PQR. A 29 cm
R

52º Solution
Q P
O (a) The angle subtended by the
diameter ‾ AB is 90°. Hence, the

side ‾ AC can be obtained by using
the Pythagoras' theorem. That is,
Solution 2 2 2
From the figure, it implies that ( ‾ AC ) + ( ‾ CB ) = ( ‾ AB )
∠PRQ = 90° (angle in a semicircle). ___________ 2
2
‾ = √ ( ‾ AB ) − ( ‾ CB )
Thus, AC
_________________
2
2
∠PQR + ∠PRQ + 52° = 180° = √ (29 cm) − (21 cm)

_____
(sum of angles in a triangle). = 400 cm
2
√

Substitute ∠PRQ to obtain,
= 20 cm
∠PQR + 90° + 52° = 180°
Therefore, ‾ AC = 20 cm.
⇒ ∠PQR = 180° − 142°

= 38°
(b) From trigonometric ratios, it
Therefore, ∠PQR = 38° . follows that
̂ BC
‾
_
cos A B C =
Example 5.15 B Mathematics for Secondary Schools
‾ A
21 cm
= _
In the following figure, ‾ AB is the 29 cm
diameter of the circle and C is a point Thus,
̂
on the circumference. If ‾ AB = 29 cm A B C = cos
−1 21
_
(
)
and ‾ BC = 21 cm, find: 29
= 43 . 6°
(a) ‾ AC (b) ∠ABC Therefore, ∠ABC = 43 . 6° .

Student\s Book Form Three 137

18/09/2025 09:59:45
MATHEMATIC F3 SB.indd 137 18/09/2025 09:59:45
MATHEMATIC F3 SB.indd 137