3.2.1 Circles II, PT3 Practice
Question 1:
Find the value of x and of y.
(b) In the diagram below, O is the centre of the circle. Find the value of

(i) x (ii) y
Solution:
(a)
∠LOM=∠LNM=yo=44o∴y=44∠NMO=∠NLO=41oxo=180o−80o−41o=59o
∠LOM=∠LNM=yo=44o∴y=44∠NMO=∠NLO=41oxo=180o−80o−41o=59o
(b)(i)
2x = 200o
x = 100o
(b)(ii)
x + y = 180o
100o + y = 180o
y = 80oQuestion 2:

Find the value of y.

Find the value of y.
Solution:
(a)
∠AOB=45o×2=90oyo=180o−90o=90o
∠AOB=45o×2=90oyo=180o−90o=90o
(b)
∠ABD=12×100o=50o∠CDE=∠ABC=85oyo+50o=85oyo=35oy=35
Question 3:
(a) In diagram below, ABCD is a circle. AEC and BED are straight lines.

Find the value of y.
(b) Diagram below shows a circle KLMN with centre O.

Find the value of x.
Solution:
(a)
∠ABE=∠ACD=45oyo=180o−45o−55oyo=80oy=80
(b)
∠OML=∠OLM=42o(xo+42o)+∠KNM=180oxo+42o+110o=180oxo+152o=180oxo=180o−152oxo=28ox=28
Question 4:
Diagram below shows a semicircle ABCD with centre O.
It is given that AB = CD.
Find the value of y.
Solution:
yo + 32o + 32o + 90o = 180o
yo = 180o – 154o
yo = 26o
y = 26
Question 5:
In diagram below, ABC and DEF are straight lines.

Find the value of x and of y.
Solution:
x+38=93 x=93−38 x=55∠DBG=38o (GB=GD) yo=∠DBG ∴y=38
In diagram below, ABC and DEF are straight lines.

Find the value of x and of y.
Solution:
x+38=93 x=93−38 x=55∠DBG=38o (GB=GD) yo=∠DBG ∴y=38