Page 141 - Mathematics_Form_3

Circles

Example 5.12
D G
̂ H
In the following figure, Q B C = b, 3n-10º 84.5º
̂
̂
B Q C = q. Find A P B in terms of q
and b.
FOR ONLINE READING ONLY
P
2n+30º
F
2m
E
A

B Solution
b From the given figure, it follows that
2 m = 84 . 5° (the exterior angle of a
D q cyclic quadrilateral).
C Q
Thus,
Solution
̂
1 _
Given A P B = p . Using ΔBCQ, it m = × 84 . 5°

2
̂
gives B C Q + b + q = 180° (sum of = 42 . 25°
angles of ΔBCQ ). In a cyclic quadrilateral DEFG, it
̂
Thus, B C Q = 180° − (b + q) . implies that
̂
̂
B A D = A B P + p (exterior angle of 3n − 10° + 2n + 30° = 180°

ΔABP ). (opposite angles in a cyclic
̂
̂
But, B C Q = B A D (exterior angle of quadrilateral).
a cyclic quadrilateral).
̂ ̂ Thus, 5n = 160°
Hence, A B P + p = 180° − (b + q) .
Mathematics for Secondary Schools Thus, b + p = 180° − (b + q ). ⇒= 32°
̂
5n + 20 ° = 180°
But, A B P = C B Q = b (vertically
opposite angles).
⇒
n
p = 180° − b − q − b
Therefore, m = 42 . 25° and n = 32° .
= 180° − 2b − q
̂
Therefore, A P B = 180° − 2b − q .
Exercise 5.7

Example 5.13
In the following figure, DEFG is a 1. Given the following figure, find the
value of x, hence the measures of
ˆ
cyclic quadrilateral. If FGH 84.5= °,
angles QRS and PTS.
find the values of m and n.

134 Student\s Book Form Three

18/09/2025 09:59:43
MATHEMATIC F3 SB.indd 134 18/09/2025 09:59:43
MATHEMATIC F3 SB.indd 134